How To Solve Kirchoff'S Law Problems?

How To Solve Kirchoff’S Law Problems?

How To Solve Kirchoff
How To Solve Physics Problems Kirchoff’s Laws problems and solutions Kirchhoff’s laws applied to circuits containing multiple branches, voltage sources, and resistors allow calculation of currents in each branch of the circuit. Start with a very simple circuit consisting of a battery and resistor.

The battery polarity is taken as indicated with positive charge Fig.31-1 leaving the positive plate. The current is as indicated by the arrow. An amount of charge or current over time starting at the negative plate of the battery is viewed as gaining energy in passing through the battery and losing this same amount of energy in the resistor.

The equation for this circuit is shown in Fig.31-1. Kirchhoff Voltage Loop Equations. The V – IR equation for this simple circuit illustrates the first Kirchhoff law: The algebraic sum of the changes in potential (energy) encountered in a complete traverse of a path is zero.

This is analogous to a conservation of energy statement. The only difficulty in applying Kirchhoff’s laws is keeping the algebraic signs correct. When confused, go back to this simple diagram of a battery and resistor and the current flowing (in the external circuit) from the positive side of the battery through the resistor to the negative side of the battery.

Write the V – IR = 0 statement establishing that when a battery is traversed in a positive direction, there is a gain in energy giving V a plus sign; and when a resistor is traversed in the direction of the current arrow, there is a loss in energy giving IR a negative sign.

Every Kirchhoff law problem you do should have this little circuit drawn in a corner of the paper to remind you of the algebraic signs and their importance.
Think of a “piece of charge” traversing the circuit gaining energy as it passes through the battery and losing energy as it passes through the resistor.

In this simple circuit the direction of the current is clear. In multiple branch circuits it is not possible to “guess” the correct directions for the currents. Make the best educated guess that you can and let the mathematics tell you whether the currents are positive or negative.

Fig.31-2 Consider the three branch circuit of Fig.31-2.
Make your best estimate of the current direction for each resistor and draw and label the current arrows.
It is not necessary to have the direction correct, it is only necessary to be consistent in applying the sign convention.
Now make two loops around this circuit, one around the left and one around the right as indicated by the closed loops.

Remember, when traversing a battery from minus to positive write + V and when traversing the resistor in the direction of the current arrow write – IR. Around the first loop

In traversing this loop start at the low, or negative, side of the 6V battery and proceed around the loop: 6 is positive
4 I 1 is negative because the resistor is traversed in the direction of the current arrow.
3 I 2 is positive because the resistor is traversed in the direction opposite to the current arrow.
7 is negative because the battery is traversed from plus to minus.
Around the second loop
Rewrite these two equations

Writing the equations this way illustrates very clearly that they cannot be solved. In mathematical language they are two equations in three unknowns. The third equation necessary to solve for the three currents comes from the next Kirchhoff law. Kirchhoff Current Junction Equations The second Kirchhoff law deals with the currents: the sum of the currents to any junction must add to zero.

This law is often forgotten.
It is, however, easily remembered via a whimsical law called the “fat wire law.” The fat wire law is very simple.
The sum of the currents to any junction must equal zero, otherwise the wire will get fat.
Applying this second Kirchhoff law to the junction just above the 3 Ω resistor Now we have three independent equations in three unknowns.

NOTE: For convenience in the discussion, individual equations in a set of equations will be designated by letter symbols. The number of significant figures used in the calculations will be kept to a minimum so as not to obscure the mathematical procedure with numbers.

Likewise, units will not be used. Several techniques for solving multiple equations in multiple unknowns will be presented. It is not necessary to know all these techniques. Pick a technique that you feel comfortable with and learn it well. At this point you may be asking why we did not take another loop to obtain a third equation.

Taking another loop in the circuit will only serve to generate another equation which is a linear combination of the first two loop equations. Take a loop around the outside of the circuit and write: This equation is just the difference of the first two loop equations and is not a linearly independent third equation.

1 4 I 1 +4 I 2 +4 I 3 =0
add (b) -4 I 1 +3 I 2 =1
to obtain ( α ) 7 I 2 – 4 I 3 =1
Combine equation ( α ) with 2/3 of equation (c) and write ( α ) and ( β )
Place I 2 = 1/3 into (c).
3(1/3)+6 I 3 =3 or I 3 = 1/3

Now place I 2 = 1/3 and I 3 = 1/3 into (a). I 1 +(1/3)-(1/3)=0 or I 1 =0 Check these current values in each of the original equations to verify that they are correct. Solving by Determinants Another method of solving these equations is with determinants.

Using determinants to solve three equations in three unknowns is explained in the Introduction, Mathematical Background.
Your calculator may have an algorithm for solving simultaneous equations using determinants (sometimes called Cramer’s rule), so this may be your method of choice.
Going back to the equations marked (a), (b), (c) set up the master determinant and expand along the first column.

Equation (c) is divided by 3 to reduce the numbers.

The determinant for I 1 is expanded along the top row
The value of
The determinant for I 2 is expanded about the top row The value for
As an exercise set up the determinant for I 3, expand along the top row, and verify the value of I 3 obtained previously.
Solving by Augmented Matrix

A third method of solving three equations in three unknowns is with an augmented matrix. Many calculators that have internal programs for solving simultaneous equations require data to be input in a manner similar to the augmented matrix form. The manipulations of the augmented matrix are similar to adding and subtracting equations to find a solution.

Before applying this method to the equations for this problem look at a general set of three equations in three unknowns.
If the first equation were multiplied by -(k/a) and added to the third equation these three equations would read What this accomplishes is to make the coefficient of x zero in the third equation.

Now multiply the first equation by -k/e and add to the second equation making the coefficient of x zero in the second equation. After this, the second equation can be multiplied by a constant and added to the third equation with the constant chosen so as to make the coefficient of y in the third equation zero.

Once this is accomplished the solutions are easily written down.
The procedure is better illustrated with a specific case, the previous problem.
These equations already have some coefficients equal to zero making the process easier.
This is usually the case with Kirchhoff law analysis Multiply the first equation by 4 and add to the second equation generating a new second equation.

This makes the coefficient of I 1 equal to zero in the second and third equation producing two equations in two unknowns. Now multiply the second equation by -1/7 and add to the third equation generating a new third equation. Now solve directly with I 3 =1/3.

Substituting this into the second equation I 2 =1/3. Putting these two values into the first equation obtains I 1 =0. The augmented matrix method uses the coefficients from the equations. The adding and subtracting of equations to generate other equations is done by multiplying and adding different rows of the matrix just as if they were equations.

As we go through the process imagine that the operations are being performed on equations. Working with the coefficients arranged in an orderly array is easier. Working the matrix so as to produce zeros starting in the lower left hand corner and then moving diagonally across the matrix is easier than working the equations.

Start with There is already a zero in the lower left corner so we need to create a zero in the -4 position.
Multiply the first row by 4 and add to the second row generating a new second row.
This is equivalent to multiplying the first equation by 4 and adding it to the second equation generating a new second equation, which is a linear combination of the first two.

This is exactly what is done in solving simultaneous equations by adding and subtracting.

Now multiply the third row by -7 and add to the second row generating a new third row.
Now go back to the equations where 18 I 3 = 6 or I 3 =1/3.
Put I 3 =1/3 into 7 I 2 -4 I 3 =1 to obtain I 2 =1/3.
Put I 2 =1/3 and I 3 =1/3 to obtain I 1 =0.

These methods of solving multiple equations in multiple unknowns require considerable number manipulation with the attendant chance for error. The augmented matrix method has the least number manipulation while adding and subtracting may be more familiar to you.

All of the techniques work.
Pick the one that fits you, and the set of equations you are solving, best and proceed carefully.
Applications The procedure for solving Kirchhoffs laws problems is the following: 1.
Draw the battery and resistor circuit in the corner of your paper to remind you of the sign convention.2.

Place current arrows next to each resistor and label them.3. Pick a junction and write the current statement.4. Draw loops and write the loop equations.5. Solve for the currents and check them in the original equations.31-1 Solve for the current in each of the resistors in the circuit shown in Fig.31-3.

Fig.31-3 Solution: Follow the procedure for Kirchhoff problems.
This is a subject that is conceptually easy but because of the large number of manipulations often hard to actually do.
Work this problem as you are following along in the book.
Draw the circuit and begin.
Draw the single battery and single resistor with current arrow and V – IR =0 equation.

Place and label the current arrows. Write the statement of the “fat wire law” for the top center of the circuit. Write a loop statement for the left side of the circuit. Write a loop statement for the right side of the circuit. Now write these three equations in a convenient form for solution.

As an exercise solve these equations and verify that the solutions are I 1 = 17/47 = 0.36, I 2 =24/47 = 0.51, and I 3 =7/47=0.15.
Check these answers in the original equations.31-2 Consider a more complicated problem as shown in Fig.31-4 and calculate the currents in the resistors.
Fig.31-4 Solution: Draw the battery and resistor circuit and V – IR equation to remind you of the algebraic sign convention.

Note the current arrows. Remember that if one of the currents comes out negative it means only that the current is opposite to the direction of your arrow.

Apply the “fat wire law” to the line across the top of the three resistors.
The loop around the top branch of the circuit is 6-8 I 1 =0.
A loop around the lower left branch is 10-10 I 4 -5 I 3 -3=0.
A loop around the lower right branch is 3+5 I 3 +6 I 2 -6=0.
Rewrite the equations in convenient form for solving.

Immediately I 1 =3/4=0.75. Work the problem and verify that the other currents are I 2 =-0.20, I 3 =0.84, and I 4 =0.28.31-3 Calculate the currents in the resistors of the circuit shown in Fig.31-5. Fig.31-5 Solution: Draw the battery and resistor circuit and V – IR equation to remind you of the algebraic sign convention.

The middle loop is 4 I 2 – 2 + 3 I 2 – 8 I 3 = 0.
The loop around the right branch is -3 + 5 I 4 + 8 I 3 = 0.
Rewriting the equations in convenient form for solving

Work the problem and verify that the currents are I 1 =0.41, I 2 =0.51, I 3 =0.19, and I 4 =0.29. Hint: First solve for I 3 using determinants, then the rest of the currents by algebra. : How To Solve Physics Problems Kirchoff’s Laws problems and solutions

How to solve Kirchhoff’s law?

Kirchhoff law – problems and solutions 1. If R 1 = 2Ω, R 2 = 4Ω, R 3 = 6Ω, determine the electric current flows in the circuit below. Known : 1 (R 1 ) = 2Ω

Resistor 2 (R 2 ) = 4Ω
Resistor 3 (R 3 ) = 6Ω
Source of emf 1 (E 1 ) = 9 V
Source of emf 2 (E 2 ) = 3 V
Wanted: Electric current (I)
Solution :
This question relates to, How to solve this problem:

First, choose the direction of the current. You can choose the opposite current or direction in the clockwise direction. Second, when the current through the resistor (R) there is a potential decrease so that V = IR signed negative. Third, if the current moves from low to high voltage (- to +) then the source of emf (E) signed positive because of the charging of energy at the emf source.

In this solution, the direction of the current is the same as the direction of clockwise rotation.
– I R 1 + E 1 – I R 2 – I R 3 – E 2 = 0
– 2 I + 9 – 4 I – 6 I – 3 = 0
– 12 I + 6 = 0
– 12 I = – 6
I = -6 / -12
I = 0.5

The electric current flows in the circuit are 0.5 A. The electric current signed positive means that the direction of the electric current is the same as the direction of clockwise rotation. If the electric current is negative then the electric current is opposite the clockwise direction.2. Determine the electric current that flows in the circuit as shown in the figure below.

Solution :
In this solution, the direction of the current is the same as the direction of clockwise rotation.
-20 – 5I -5I – 12 – 10I = 0
-32 – 20I = 0
-32 = 20I
I = -32 / 20
I = -1.6 A

Because the electric current is negative, the direction of the electric current is actually opposite to the clockwise direction. The direction of electric current is not the same as estimation.3. Determine the electric current that flows in the circuit as shown in the figure below.

Solution :
In this solution the direction of current is same as the direction of clockwise rotation.
– I – 6I + 12 – 2I + 12 = 0
-9I + 24 = 0
-9I = -24
I = 24 / 9
I = 8 / 3 A

4. An electric circuit consists of four resistors, R 1 = 12 Ohm, R 2 = 12 Ohm, R 3 = 3 Ohm and R 4 = 6 Ohm, are connected with source of emf E 1 = 6 Volt, E 2 = 12 Volt. Determine the electric current flows in the circuit as shown in figure below.

Known :
Resistor 1 (R 1 ) = 12 Ω
Resistor 2 (R 2 ) = 12 Ω
Resistor 3 (R 3 ) = 3 Ω
Resistor 4 (R 4 ) = 6 Ω
Source of emf 1 (E 1 ) = 6 Volt
Source of emf 2 (E 2 ) = 12 Volt
Wanted : The electric current flows in the circuit (I)
Solution :
Resistor 1 ( R 1 ) and resistor 2 ( R 2 ) are connected in parallel. The equivalent resistor :
1/R 12 = 1/R 1 + 1/R 2 = 1/12 + 1/12 = 2/12
R 12 = 12/2 = 6 Ω
In this solution, the direction of current is same as the direction of clockwise rotation.
– I R 12 – E 1 – I R 3 – I R 4 + E 2 = 0
– 6 I – 6 – 3I – 6I + 12 = 0
– 6I – 3I – 6I = 6 -12
– 15I = – 6
I = -6/-15
I = 2/5 A

5. Determine the electric current that flows in circuit as shown in figure below.

Known :
Resistor 1 (R 1 ) = 10 Ω
Resistor 2 (R 2 ) = 6 Ω

Resistor 3 (R 3 ) = 5 Ω
Resistor 4 (R 4 ) = 20 Ω
Source of emf 1 (E 1 ) = 8 Volt
Source of emf 2 (E 2 ) = 12 Volt
Wanted : The electric current that flows in circuit
Solution :
Resistor 3 ( R 3 ) and resistor 4 ( R 4 ) are connected in parallel. The equivalent resistor :
1/R 34 = 1/R 3 + 1/R 4 = 1/5 + 1/20 = 4/20 + 1/20 = 5/20
R 34 = 20/5 = 4 Ω
In this solution, the direction of current is same as the direction of clockwise rotation.
– I R 1 – I R 2 – E 1 – I R 34 + E 2 = 0
– 10I – 6I – 8 – 4I + 12 = 0
– 10I – 6I – 4I = 8 – 12
– 20 I = – 4
I = -4/-20
I = 1/5 A
I = 0.2 A

6. Determine the electric current that flows in circuit as shown in figure below.

Known :
Resistor 1 (R 1 ) = 1 Ω
Resistor 2 (R 2 ) = 6 Ω
Resistor 3 (R 3 ) = 6 Ω
Resistor 4 (R 4 ) = 4 Ω
Source of emf 1 (E 1 ) = 12 Volt
Source of emf 2 (E 2 ) = 6 Volt
Wanted : The electric current that flows in circuit
Solution :

Resistor 1 ( R 1 ) and resistor 2 ( R 2 ) are connected in parallel. The equivalent resistor :
1/R 12 = 1/R 1 + 1/R 2 = 1/1 + 1/6 = 6/6 + 1/6 = 7/6
R 12 = 6/7 Ω
The direction of current is same as the direction of clockwise rotation.
E 1 – I R 12 – E 2 – I R 4 – I R 3 = 0
12 – (6/7)I – 6 – 4I – 6I = 0
12 – 6 – (6/7)I – 4I – 6I = 0
6 – (6/7)I – 10I = 0
6 = (6/7)I + 10I
6 = (6/7)I + (70/7)I
6 = (76/7)I
(6)(7) = 76I
42 = 76I
I = 42/76
I = 0.5 A

: Kirchhoff law – problems and solutions

What is Kirchhoff’s voltage law (KVL)?

We have gone over Kirchhoff’s Current Law (KCL) in a previous tutorial and Kirchhoff’s Voltage Law (KVL) is very similar but focused on the voltage in a circuit, not the current. Kirchhoff’s Voltage law states that the sum of the voltages in a closed loop will equal zero. Using this concept, much as how we can use nodal analysis with KCL, we can use mesh analysis because of KVL. While a mesh is basically any loop within a circuit, for mesh analysis, we will need to define meshes that don’t enclose any other meshes. You can see that if we make a loop around the ‘outside’ of the entire circuit, technically it is a mesh because the loop can be completed. However, for purposes of analysis, we need to break it into three different meshes. So let’s go over the steps of how to solve a circuit using mesh analysis before jumping into a few examples.

Take your time, breathe, and assess the problem. Write down what info you’ve been given and any intuitive insights you have.Assign mesh currents to all of the meshes. There should be one current assigned per mesh. You need to choose which direction your current is flowing – this is semi-arbitrary because as long as you do your math right, it doesn’t really matter. But in most cases, people assume a clockwise current direction.Apply KVL to each of the meshes, using Ohm’s Law to show the voltages in terms of the current.Solve the simultaneous equations (like we did with KCL) to find the actual values.Sanity check. Take a moment to review what you’ve done and see if the numbers make sense and are internally consistent.

We’ll go over some examples now and frankly, after these examples, the only real additions and changes will be complications that make the math more difficult. The problems shouldn’t get much harder conceptually but the math can get significantly harder.

What is Kirchoff’s Law of Conservation of energy?

Kirchoff’s Law with Example Problems and Calculations

KIRCHOFF’S LAW
Kirchoff’s First Law – The Current Law, (KCL)
“The total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node “.
In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero,
I(exiting) + I(entering) = 0.
This idea by Kirchoff is known as the Conservation of Charge,

How To Solve Kirchoff

Here, the 3 currents entering the node, I 1, I 2, I 3 are all positive in value and the 2 currents leaving the node, I4 and I5 are negative in value.
Then this means we can also rewrite the equation as; I 1 + I 2 + I 3 – I 4 – I 5 = 0
Kirchoff’s Second Law – The Voltage Law, (KVL)

” In any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop ” which is also equal to zero. In other words the algebraic sum of all voltages within the loop must be equal to zero. This idea by Kirchoff is known as the Conservation of Energy,

Starting at any point in the loop continue in the same direction noting the direction of all the voltage drops, either positive or negative, and returning back to the same starting point. It is important to maintain the same direction either clockwise or anti-clockwise or the final voltage sum will not be equal to zero.

We can use Kirchoff’s voltage law when analyzing series circuits. How To Solve Kirchoff Problem 1: A current of 0.5 A is flowing through the resistance of 10Ω.Findthepotential difference between its ends. Solution: Current I = 0.5A.

Resistance R = 10Ω
Potential difference V =?
V = IR
= 0.5 × 10
= 5V.
Problem: 2

A supply voltage of 220V is applied to a resistor100.FindΩthe current flowing through it.

Solution:
Voltage V = 220V Resistance R = 100Ω Current I = V/ R
= 2 2 0 /100

= 2.2 A.

Problem: 3
Calculate the resistance of the conductor if a current of 2A flows through it when the potential difference across its ends is 6V.
Solution:

Current I = 2A. Potential difference = V = 6. Resistance R = V/I

= 6 /2
= 3 ohm.
Problem: 4
Calculate the current and resistance of a 100 W, 200V electric bulb.
Solution:
Power, P = 100W
Voltage, V = 200V Power p = VI
Current I = P/V
= 100/200
= 0.5A
Resistance R = V /I
= 200/0.5
= 400W.
Problem: 5
Calculate the power rating of the heater coil when used on 220V supply taking 5 Amps.
Solution:
Voltage, V = 220V Current, I = 5A, Power, P = VI
= 220 × 5
= 1100W

= 1.1 KW. Problem: 6 A circuit is made of 0.4wire,Ωa 150bulbΩand a rheostat120 connected Ω in series. Determine the total resistance of the resistance of the circuit.

Solution:
Resistance of the wire = 0.4ResistanceΩ of bulb = 1 5 0 Ω Resistance of rheostat = 120Ω
In series,

Total resistance, R = 0.4 + 150 +120 = 270.4Ω Problem : 7 Three resistances of values 2Ω,3Ωconnectedandin series5Ω acrossare20 V,D.C supply,Calculate (a) equivalent resistance of the circuit (b) the total current of the circuit (c) the voltage drop across each resistor and (d) the power dissipated in each resistor.

Solution:
Total resistance R = R1 + R 2+ R3.
= 2 +3+5 = 10Ω
Voltage = 20V
Total current I = V/R = 20/10 = 2A.
Voltage drop across 2Ωresistor V1 = I R 1
= 2× 2 = 4 volts.
Voltage drop across 3Ωresistor V 2 = IR 2
= 2 × 3 = 6 volts.
Voltage drop across 5Ωresistor V 3 = I R 3
= 2 ×5 = 10 volts.
Power dissipated in 2Ωresistor is P 1 = I 2 R 1
= 22 × 2 = 8 watts.
Power dissipated in 3 resistor is P 2 = I 2 R 2,
= 22 × 3 = 12 watts.
Power dissipated in 5 resistor is P 3 = I 2 R 3
= 22 × 5 = 20 watts.
Problem: 8

A lamp can work on a 50 volt mains taking 2 amps. What value of the resistance must be connected in series with it so that it can be operated from 200 volt mains giving the same power.

Solution:
Lamp voltage, V = 50V Current, I = 2 amps.
Resistance of the lamp = V/I = 50/25 = 25 Ω
Resistance connected in series with lamp = r.
Supply voltage = 200 volt.
Circuit current I = 2A
Total resistance Rt= V/I = 200/2 = 100Ω
R t = R + r 100 = 25 + r
r = 75Ω
Problem: 9
Find the current flowing in the 40Ω Resistor,

How To Solve Kirchoff

Solution:
The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.
Using Kirchoff’s Current Law, KCL the equations are given as;
At node A: I 1 + I 2 = I 3
At node B: I 3 = I 1 + I 2
Using Kirchoff’s Voltage Law, KVL the equations are given as;
Loop 1 is given as: 10 = R 1 x I 1 + R 3 x I 3 = 10I 1 + 40I 3
Loop 2 is given as: 20 = R 2 x I 2 + R 3 x I 3 = 20I 2 + 40I 3
Loop 3 is given as: 10 – 20 = 10I 1 – 20I 2
As I3 is the sum of I 1 + I 2 we can rewrite the equations as;
Eq. No 1: 10 = 10I 1 + 40(I 1 + I 2 ) = 50I 1 + 40I 2
Eq.No 2: 20 = 20I 1 + 40(I 1 + I 2 ) = 40I 1 + 60I 2
We now have two “Simultaneous Equations” that can be reduced to give us the value of both I 1 and
I 2
Substitution of I 1 in terms of I 2 gives us the value of I 1 as -0.143 Amps
Substitution of I 2 in terms of I 1 gives us the value of I 2 as +0.429 Amps
As: I 3 = I 1 + I 2

The current flowing in resistor R 3 is given as: -0.143 + 0.429 = 0.286 Amps and the voltage across the resistor R 3 is given as : 0.286 x 40 = 11.44 volts

Problem: 10
Find the current in a circuit using Kirchhoff’s voltage law

How do you calculate loop 3 using Kirchoff’s law?

Kirchoff’s Law with Example Problems and Calculations

KIRCHOFF’S LAW
Kirchoff’s First Law – The Current Law, (KCL)
“The total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node “.
In other words the algebraic sum of ALL the currents entering and leaving a node must be equal to zero,
I(exiting) + I(entering) = 0.
This idea by Kirchoff is known as the Conservation of Charge,

How To Solve Kirchoff

Here, the 3 currents entering the node, I 1, I 2, I 3 are all positive in value and the 2 currents leaving the node, I4 and I5 are negative in value.
Then this means we can also rewrite the equation as; I 1 + I 2 + I 3 – I 4 – I 5 = 0
Kirchoff’s Second Law – The Voltage Law, (KVL)

Starting at any point in the loop continue in the same direction noting the direction of all the voltage drops, either positive or negative, and returning back to the same starting point. It is important to maintain the same direction either clockwise or anti-clockwise or the final voltage sum will not be equal to zero.

Resistance R = 10Ω
Potential difference V =?
V = IR
= 0.5 × 10
= 5V.
Problem: 2

A supply voltage of 220V is applied to a resistor100.FindΩthe current flowing through it.

Solution:
Voltage V = 220V Resistance R = 100Ω Current I = V/ R
= 2 2 0 /100

= 2.2 A.

Problem: 3
Calculate the resistance of the conductor if a current of 2A flows through it when the potential difference across its ends is 6V.
Solution:

Current I = 2A. Potential difference = V = 6. Resistance R = V/I

= 6 /2
= 3 ohm.
Problem: 4
Calculate the current and resistance of a 100 W, 200V electric bulb.
Solution:
Power, P = 100W
Voltage, V = 200V Power p = VI
Current I = P/V
= 100/200
= 0.5A
Resistance R = V /I
= 200/0.5
= 400W.
Problem: 5
Calculate the power rating of the heater coil when used on 220V supply taking 5 Amps.
Solution:
Voltage, V = 220V Current, I = 5A, Power, P = VI
= 220 × 5
= 1100W

= 1.1 KW. Problem: 6 A circuit is made of 0.4wire,Ωa 150bulbΩand a rheostat120 connected Ω in series. Determine the total resistance of the resistance of the circuit.

Solution:
Resistance of the wire = 0.4ResistanceΩ of bulb = 1 5 0 Ω Resistance of rheostat = 120Ω
In series,

Solution:
Total resistance R = R1 + R 2+ R3.
= 2 +3+5 = 10Ω
Voltage = 20V
Total current I = V/R = 20/10 = 2A.
Voltage drop across 2Ωresistor V1 = I R 1
= 2× 2 = 4 volts.
Voltage drop across 3Ωresistor V 2 = IR 2
= 2 × 3 = 6 volts.
Voltage drop across 5Ωresistor V 3 = I R 3
= 2 ×5 = 10 volts.
Power dissipated in 2Ωresistor is P 1 = I 2 R 1
= 22 × 2 = 8 watts.
Power dissipated in 3 resistor is P 2 = I 2 R 2,
= 22 × 3 = 12 watts.
Power dissipated in 5 resistor is P 3 = I 2 R 3
= 22 × 5 = 20 watts.
Problem: 8

A lamp can work on a 50 volt mains taking 2 amps. What value of the resistance must be connected in series with it so that it can be operated from 200 volt mains giving the same power.

Solution:
Lamp voltage, V = 50V Current, I = 2 amps.
Resistance of the lamp = V/I = 50/25 = 25 Ω
Resistance connected in series with lamp = r.
Supply voltage = 200 volt.
Circuit current I = 2A
Total resistance Rt= V/I = 200/2 = 100Ω
R t = R + r 100 = 25 + r
r = 75Ω
Problem: 9
Find the current flowing in the 40Ω Resistor,

How To Solve Kirchoff

Solution:
The circuit has 3 branches, 2 nodes (A and B) and 2 independent loops.
Using Kirchoff’s Current Law, KCL the equations are given as;
At node A: I 1 + I 2 = I 3
At node B: I 3 = I 1 + I 2
Using Kirchoff’s Voltage Law, KVL the equations are given as;
Loop 1 is given as: 10 = R 1 x I 1 + R 3 x I 3 = 10I 1 + 40I 3
Loop 2 is given as: 20 = R 2 x I 2 + R 3 x I 3 = 20I 2 + 40I 3
Loop 3 is given as: 10 – 20 = 10I 1 – 20I 2
As I3 is the sum of I 1 + I 2 we can rewrite the equations as;
Eq. No 1: 10 = 10I 1 + 40(I 1 + I 2 ) = 50I 1 + 40I 2
Eq.No 2: 20 = 20I 1 + 40(I 1 + I 2 ) = 40I 1 + 60I 2
We now have two “Simultaneous Equations” that can be reduced to give us the value of both I 1 and
I 2
Substitution of I 1 in terms of I 2 gives us the value of I 1 as -0.143 Amps
Substitution of I 2 in terms of I 1 gives us the value of I 2 as +0.429 Amps
As: I 3 = I 1 + I 2

The current flowing in resistor R 3 is given as: -0.143 + 0.429 = 0.286 Amps and the voltage across the resistor R 3 is given as : 0.286 x 40 = 11.44 volts

Problem: 10
Find the current in a circuit using Kirchhoff’s voltage law

How to solve Kirchhoff’s law?

Kirchhoff law – problems and solutions 1. If R 1 = 2Ω, R 2 = 4Ω, R 3 = 6Ω, determine the electric current flows in the circuit below. Known : 1 (R 1 ) = 2Ω

Resistor 2 (R 2 ) = 4Ω
Resistor 3 (R 3 ) = 6Ω
Source of emf 1 (E 1 ) = 9 V
Source of emf 2 (E 2 ) = 3 V
Wanted: Electric current (I)
Solution :
This question relates to, How to solve this problem:

In this solution, the direction of the current is the same as the direction of clockwise rotation.
– I R 1 + E 1 – I R 2 – I R 3 – E 2 = 0
– 2 I + 9 – 4 I – 6 I – 3 = 0
– 12 I + 6 = 0
– 12 I = – 6
I = -6 / -12
I = 0.5

Solution :
In this solution, the direction of the current is the same as the direction of clockwise rotation.
-20 – 5I -5I – 12 – 10I = 0
-32 – 20I = 0
-32 = 20I
I = -32 / 20
I = -1.6 A

Solution :
In this solution the direction of current is same as the direction of clockwise rotation.
– I – 6I + 12 – 2I + 12 = 0
-9I + 24 = 0
-9I = -24
I = 24 / 9
I = 8 / 3 A

Known :
Resistor 1 (R 1 ) = 12 Ω
Resistor 2 (R 2 ) = 12 Ω
Resistor 3 (R 3 ) = 3 Ω
Resistor 4 (R 4 ) = 6 Ω
Source of emf 1 (E 1 ) = 6 Volt
Source of emf 2 (E 2 ) = 12 Volt
Wanted : The electric current flows in the circuit (I)
Solution :
Resistor 1 ( R 1 ) and resistor 2 ( R 2 ) are connected in parallel. The equivalent resistor :
1/R 12 = 1/R 1 + 1/R 2 = 1/12 + 1/12 = 2/12
R 12 = 12/2 = 6 Ω
In this solution, the direction of current is same as the direction of clockwise rotation.
– I R 12 – E 1 – I R 3 – I R 4 + E 2 = 0
– 6 I – 6 – 3I – 6I + 12 = 0
– 6I – 3I – 6I = 6 -12
– 15I = – 6
I = -6/-15
I = 2/5 A

5. Determine the electric current that flows in circuit as shown in figure below.

Known :
Resistor 1 (R 1 ) = 10 Ω
Resistor 2 (R 2 ) = 6 Ω

Resistor 3 (R 3 ) = 5 Ω
Resistor 4 (R 4 ) = 20 Ω
Source of emf 1 (E 1 ) = 8 Volt
Source of emf 2 (E 2 ) = 12 Volt
Wanted : The electric current that flows in circuit
Solution :
Resistor 3 ( R 3 ) and resistor 4 ( R 4 ) are connected in parallel. The equivalent resistor :
1/R 34 = 1/R 3 + 1/R 4 = 1/5 + 1/20 = 4/20 + 1/20 = 5/20
R 34 = 20/5 = 4 Ω
In this solution, the direction of current is same as the direction of clockwise rotation.
– I R 1 – I R 2 – E 1 – I R 34 + E 2 = 0
– 10I – 6I – 8 – 4I + 12 = 0
– 10I – 6I – 4I = 8 – 12
– 20 I = – 4
I = -4/-20
I = 1/5 A
I = 0.2 A

6. Determine the electric current that flows in circuit as shown in figure below.

Known :
Resistor 1 (R 1 ) = 1 Ω
Resistor 2 (R 2 ) = 6 Ω
Resistor 3 (R 3 ) = 6 Ω
Resistor 4 (R 4 ) = 4 Ω
Source of emf 1 (E 1 ) = 12 Volt
Source of emf 2 (E 2 ) = 6 Volt
Wanted : The electric current that flows in circuit
Solution :

Resistor 1 ( R 1 ) and resistor 2 ( R 2 ) are connected in parallel. The equivalent resistor :
1/R 12 = 1/R 1 + 1/R 2 = 1/1 + 1/6 = 6/6 + 1/6 = 7/6
R 12 = 6/7 Ω
The direction of current is same as the direction of clockwise rotation.
E 1 – I R 12 – E 2 – I R 4 – I R 3 = 0
12 – (6/7)I – 6 – 4I – 6I = 0
12 – 6 – (6/7)I – 4I – 6I = 0
6 – (6/7)I – 10I = 0
6 = (6/7)I + 10I
6 = (6/7)I + (70/7)I
6 = (76/7)I
(6)(7) = 76I
42 = 76I
I = 42/76
I = 0.5 A

: Kirchhoff law – problems and solutions

What is Kirchhoff’s voltage law (KVL)?

What is Kirchhoff’s loop rule?

FAQs on Kirchhoff’s Laws – Q.1. What is the Junction and loop Rule? Ans: The junction rule is also known as Kirchhoff’s Current Law KCL and it states that at any junction the sum of the entering currents is equal to the sum of the leaving currents.Kirchhoff’s Loop Rule also known as Kirchhoff’s Voltage Law KVL and it states that the sum of the voltage differences around the loop must be equal to zero.Q.2.

What is Node Voltage? Ans: When we use the term node voltage, we are referring to the potential difference between two nodes of a circuit. We select one of the nodes in the given circuit as a reference node. All the voltages of other nodes are measured concerning this one reference node.Q.3. What is the importance of Kirchhoff’s law in daily life? Ans: Kirchhoff’s laws can be used to determine the values of unknown values like current, voltage in the circuit.

These laws can be applied in any circuit (with some limitations), and useful to find the unknown values in complex circuits and networks. It helps in knowing the energy transfer in different parts of the circuit.Q.4. Does Kirchhoff’s law fail at high frequency? Ans: Yes, Kirchhoff’s laws fail at high frequency, because both the law \( }}\) and \( }}\) are not suitable for \( }}\) circuits of high frequencies.

How to solve a circuit?

Sample Problem 1 – Now that we know that, according to KCL, everything that goes into a node must also come out, we can start learning about nodal analysis. Looking at the figure above, you can see that none of the resistors are in series or parallel, so this can’t be simplified. Also note that there are no voltage sources, just current sources, which makes KCL simpler. Step 1) The first step I always take when solving a circuit is to review what’s been given, what I need, and to take a moment and breathe.

Sometimes a circuit can be overwhelming but just taking your time and going through this process will inevitably simplify the problem.
In this case, we have three resistors and we know their values.
We’ve been given two current sources and we know their values.
So, there are only two nodes (N 1 and N 2 ) whose voltage we need and one branch that we need to know the current through (R 3 ).

Since we identified we need to know the current through R 3, let’s put an arrow pointing down next to R 3 and label it as I 3, Step 2) Now that we’ve written down everything we know and what we need to know, let’s choose a reference ground, if it hasn’t already been done. This is also semi-arbitrary. You can choose any node you want in a circuit, in this case, either N 1 or N 2 would work and the math would work out. Step 3) Write out the equations for the current through different branches. In this case, we already know the current through R 1 and R 2 because it’s I 1 (which is 1A) and I 2 (which is 2A). However, we don’t know the current through R 3 or I 3 yet. So, we can define that as the voltage at N 1 minus N 2 – our reference voltage in this case – over the resistance. In other words: Step 4) Using the equations you’ve created (only the one in this case), put together the equations for the current into and out of each node. So, at N 1 we know that we have I 1 and I 2 going in and I 3 going out. Again, we’re assuming all of this (and it’s a very reasonable assumption in this case) and if our assumptions are wrong, the way that will be shown is with negative currents and voltages. As we have only one unknown, we only have this single equation. However, when there are more unknowns, there will be more equations. To solve for the unknowns, you need at least as many equations as unknowns. For example, if you need two voltages, you need two equations to be able to get real values for those two voltages. We can even use Ohm’s Law to see how much voltage is created by the current sources to supply their rated currents. For example, with R 1, to get 1A flowing through a 10Ω resistor, we can see that V = IR is V = 1*10 so the voltage across R 1 must be 10V. However, since N 1 is 90V, that means that the voltage on the other side of the resistor must be 100V to get that 10V drop across R 1, I’d recommend trying to figure out what the voltage is across R 2 and what the voltage out of that current source is. So, in summary, the steps are:

Review what you have, label everything you can, establish reference flows. Breathe, take your time, don’t panic.Choose a reference ground if necessary.Start writing equations for the current through different branches.Using the equations for the current, put together equations for the current in an out of each node.Solve the equations. For multiple equations, either use linear algebra (matrices) or solve for one variable and insert that variable in the next equation, until you find real numbers and then go back and put those real numbers in for each value.