Kirchhoff's and Ohm's laws are used to solve DC circuits problems.

There are 3 examples; solve in the order that they are presented; this will make it easier to fully understand them.

__Problem 1__

Find current \( i \), voltages \( V_{R_1} \) and \( V_{R_2} \) in the ciruit below given that the voltage source \( e = 20 \) Volts, the resistances \( R_1 = 100 \; \Omega \) and \( R_2 = 300 \; \Omega \).

Solution to Problem 1

Apply Kirchhoff's law of voltage to the closed loop in the circle and write the equation

\( e - V_{R_1} - V_{R_2} = 0 \) (1)

Use Ohm's law to write

\( V_{R_1} = i R_1 \) and \( V_{R_2} = i R_2 \)

Substitute \( V_{R_1} \) and \( V_{R_2} \) by their expression in equation (1)

\( e - i R_1 - i R_2 = 0 \)

Rearrange the above so that all term containing \( i \) are on one side

\( i R_1 + i R_2 = e \)

Factor \( i \) out

\( i ( R_1 + R_2) = e \)

Solve for \( i \)

\( i = \dfrac{e}{R_1 + R_2} \)

Substitute known quantities

\( i = \dfrac{20}{100 + 300} = 0.05 \) A

Calculate \( V_{R_1} \) and \( V_{R_2} \) using Ohm's law

\( V_{R_1} = i R_1 = 0.05 \times 100 = 5 \) V

\( V_{R_2} = i R_2 = 0.05 \times 300 = 15\) V

__Problem 2__

Given the voltage sources \( e_1 = 20 \) V, \( e_2 = 5 \) V, and the resistances \( R_1 = 100 \; \Omega \) , \( R_2 = 300 \; \Omega \) and \( R_3 = 50 \; \Omega \), find all currents through and voltages across the resistors in the circuit.

Solution to Problem 2

Kirchhoff's Law of voltage for all three loops \( L_1 \) , \( L_2 \) and \( L_3 \) gives

Loop \( L_1 \): \( e_1 - V_{R_1} - V_{R_2} = 0 \) (1)

Loop \( L_2 \): \( V_{R_2} + e_2 - V_{R_3 } = 0 \) (2)

Loop \( L_3 \): \( e_1 - V_{R_1} + e_2 - V_{R_3 } = 0 \) (3)

Note that if we add equations (1) and (2), we obtain equation (3). Hence any 2 equations from the 3 can be used to solve the given problems. The third one does not give any extra information.

We have 3 unknowns, we therefore need another equation independent from the above equation.

Kirchhoff's Law of current applied at node A gives

\( i_1 = i_2 + i_3 \) (4)

So we select equations (1), (2) and (4) to write the system of equations

\( e_1 - V_{R_1} - V_{R_2} = 0 \) (5)

\( V_{R_2} + e_2 - V_{R_3 } = 0 \) (6)

\( i_1 = i_2 + i_3 \) (7)

Use Ohm's law to rewrite \( V_{R_1} \) and \( V_{R_2} \) as follows

\( V_{R_1} = R_1 i_1 \) and \( V_{R_2} = R_2 i_2\)

Substitute the above in equations (5) and (6) and rewrite the system of equations (5), (6) and (7) as

\( R_1 i_1 + R_2 i_2 = e_1 \) (8)

\( R_2 i_2 - R_3 i_3 = - e_2 \) (9)

\( i_1 = i_2 + i_3 \) (10)

Substitute the known quantities by their numerical values to obtain a linear system system of equations with three unknowns \( i_1 \), \( i_2\) and \( i_3 \) .

\( 100 i_1 + 300 i_2 = 20 \) (11)

\( 300 i_2 - 50 i_3 = - 5 \) (12)

\( i_1 = i_2 + i_3 \) (13)

There are many ways to solve the above system.

One way is to use equation (XIII) and substitute \( i_1 \) by \( i_2 + i_3 \) in equations (XI) and (XII) to obtain a system with two unknowns

\( 100 (i_2 + i_3) + 300 i_2 = 20 \) (14)

\( 300 i_2 - 50 i_3 = - 5 \) (15)

Rearrange to rewrite the above system as

\( 400 i_2 + 100 i_3 = 20 \) (16)

\( 300 i_2 - 50 i_3 = - 5 \) (17)

Multiply all terms of equation (17) above by 2 and rewrite the above system as

\( 400 i_2 + 100 i_3 = 20 \) (16)

\( 600 i_2 - 100 i_3 = - 10 \) (17)

Add side by side equations (16) and (17) to obtain one equation in one variable

\( 1000 i_2 = 10 \)

Solve for \( i_2 \)

\( i_2 = 10/1000 = 0.01 \) A

Substitute \( i_2 \) by \( 0.01 \) in equation (16) and solve for \( i_3 \)

\( 400 \times 0.01 + 100 i_3 = 20 \)

Solve for \( i_3 \)

\( i_3 = 0.16 \) A

Use equation (13) to solve for \(i_1\)

\( i_1 = i_2 + i_3 = 0.01 + 0.16 = 0.17\) A

We now calculate the voltages across the resistors

\( V_{R_1} = R_1 I_1 = 100 \times 0.17 = 17 \) V

\( V_{R_2 } = R_2 I_2 = 300 \times 0.01 = 3 \) V

\( V_{R_3 } = R_3 I_3 = 50 \times 0.16 = 8\) V

__Problem 3__

Find all currents through and voltages across the resistors in the circuit below, given the voltage sources \( e_1 = 20 \) V, \( e_2 = 5 \) V, and the resistances \( R_1 = 100 \; \Omega \) , \( R_2 = 120 \; \Omega \), \( R_3 = 60 \; \Omega \), \( R_4 = 40 \; \Omega \), \( R_5 = 240\; \Omega \) and \( R_6 = 80 \; \Omega \).

Solution to Problem 3

All the details of the solution are presented and in order to make this presentation clear and easy to understand, the solution of this example has 5 parts.

Part 1: Simplify the circuit by grouping resistors

If we apply Kirchhoff's law to the closed loops and nodes in the given circuit above, we will end up with a large number of equations to solve.

However, a quick analysis of the given circuit shows that some of the resistors are parallel and series configurations as shown below.

Resistors \( R_2 \) and \( R_3 \) are in parallel and their equivalent resistance is \( R'_2 \) as shown in the circuit below.

\( R_5\) and \( R_6 \) are in parallel and their equivalent resistance is in series with \( R_4 \) and their overall equivalent resistance is \( R'_3 \) as shown in the circuit below.

Use the formula for resistors in parallel to write

\( \dfrac{1}{R'_2} = \dfrac{1}{R_2} + \dfrac{1}{R_3} \)

Solve the above for \( R'_2 \) to obtain

\( R'_2 = \dfrac{R_2 \cdot R_3}{R_2 + R_3} \)

Substitute the knwon quantities to obtain

\( R'_2 = \dfrac{120 \cdot 60}{120 + 60} = 40 \; \Omega \)

Use formula for resistors in parallel to \( R_5\) and \( R_6 \) and add it to \( R_4 \) to write \( R'_3 \) as

\( R'_3 = R_4 + \dfrac{R_5 \cdot R_6}{R_5 + R_6} \)

Substitute the knwon quantities to obtain

\( R'_3 = 40 + \dfrac{240 \cdot 80}{240 + 80} = 100\)

Part 2: Calculate \( i_1 \), \( i_2 \) and \( i_3 \) using Kirchhoff's law

We now have a much simpler circuit to solve as shown below.

There are 3 unkwons \( i_1 \), \( i_2 \) and \( i_3 \) and we therefore need 3 equations.

Use Kirchhoff's law of current at node A to write

\( i_1 = i_2 + i_3 \) (1)

Use Use Kirchhoff's law of voltage for the loop on the left

\( e_1 - V_{R_1} - V_{R'_2} + e_2 = 0 \) (2)

Use Use Kirchhoff's law of voltage for the loop on the right

\( - e_2 + V_{R'_2} - V_{R'_3} = 0 \) (3)

Use Ohm's law to write

\( V_{R_1} = R_1 i_1 \)

\( V_{R'_2} = R'_2 i_2 \)

\( V_{R'_3} = R'_3 i_3 \)

Substitute the above voltages in equations (2) and (3) and include equation (1) to rewrite the system of 3 equations with 3 unknowns

\( e_1 - R_1 i_1 - R'_2 i_2 + e_2 = 0 \)

\( - e_2 + R'_2 i_2 - R'_3 i_3 = 0 \)

\( i_1 = i_2 + i_3 \)

Substitute the knwon quantities, simplify and rewrite the above system in the form.

\( 100 i_1 + 40 i_2 = 25\) (4)

\( 40 i_2 - 100 i_3 = 5 \) (5)

\( i_1 = i_2 + i_3 \) (6)

There are many ways to solve systems of linear equations. Let us use the method of substitution.

Substitute \( i_1 \) by \( i_2 + i_3 \) in equations (4) and rewrite equations (4) and (5) as follows

\( 100 (i_2 + i_3) + 40 i_2 = 25\) (7)

\( 40 i_2 - 100 i_3 = 5 \) (8)

Simplify and group

\( 140 i_2 + 100 i_3 = 25\)

\( 40 i_2 - 100 i_3 = 5 \)

Add the above equations to eliminate \( i_3 \)

\( 180 i_2 = 30 \)

Solve for \( i_2 \)

\( i_2 = 30/180 = 1/6 \) A

Use equation (7) (or (8) ) to find \( i_3 \)

\( 100 i_3 = 25 - 140 i_2 \)

\( i_3 = 1/60 \) A

Use equation (6) to find \( i_1 \)

\( i_1 = i_2 + i_3 = 1/6 + 1/60 = 11/60\)

Part 3: Calculate currents in resistors \( R_2 \) and \( R_3 \)

Apply Kirchhoff's law of current

\( i_2'+i_2'' = i_2 = 1/6 \)

Kirchhoff's law of voltage to the closed loop.

\( R_2 i_2' - R_3 i_2'' = 0 \)

Substitute known quantities in the above equation

\( 120 i_2' - 60 i_2'' = 0 \)

Solve the system of 2 equations 2 unknowns

\( i_2'+i_2'' = 1/6 \)

\( 120 i_2' - 60 i_2'' = 0 \)

to obtain

\( i_2' = 1/18 \) A

and \( i_2'' = 1/9 \) A

Part 4: Calculate currents in resistors \( R_5 \) and \( R_6 \)

We now calculate the currents through resistors \( R_5 \) and \( R_6 \) using the circuit below

Apply Kirchhoff's law of current

\( i_3'+i_3'' = i_3 = 1/60 \)

Kirchhoff's law of voltage to the closed loop.

\( R_5 i_3' - R_6 i_3'' = 0 \)

Substitute known quantities in the above equation

\( 240 i_2' - 80 i_2'' = 0 \)

Solve the system of 2 equations 2 unknowns

\( i_3'+i_3'' = 1/60 \)

\( 240 i_2' - 80 i_2'' = 0 \)

to obtain

\( i_3' = 1/240 \)A

and \( i_3''' = 1/80 \) A

Part 5: Calculate voltages in all resistors

We now list all resistors and the currecnts through each one of them; and then calculate the voltages across each one of them.

\( R_1 = 100 \; \Omega \) , current \( i_1 = 11/60 \) A voltage : \( V_{R_1} = 100 \times 11/60 = 18.33 \) V

\( R_2 = 120 \; \Omega \) , current \( i_2' = 1/18 \) A voltage : \( V_{R_2} = 120 \times 1/18 = 6.67 \) V

\( R_3 = 60 \; \Omega \) , current \( i_2'' = 1/9 \)A voltage : \( V_{R_3} = 60 \times 1/9 = 6.67 \) V

\( R_4 = 40 \; \Omega \) , current \( i_3 = 1/60 \) A voltage : \( V_{R_4} = 40 \times 1/60 = 0.67 \) V

\( R_5 = 240\; \Omega \) , current \( i_3' = 1/240 \)A voltage : \( V_{R_5} = 240 \times 1/240 = 1 \) V

\( R_6 = 80 \; \Omega \) , current \( i_2'' = 1/80 \)A voltage : \( V_{R_5} = 80 \times 1/80 = 1 \) V

## More References and Links

Kirchhoff's Law of Circuits with ExamplesOhm's Law of Circuits with Examples

Series and Parallel Resistors