Solve DC Circuits Problems

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