A portion of an electric circuit is depicted in the figure. The electric potentials at points $a$ , $b$ , and $c$ are given as $30 \text{ V}$ , $12 \text{ V}$ , and $2 \text{ V}$ , respectively. Determine the currents flowing through each of the three resistors.

Question

A portion of an electric circuit is depicted in the figure. The electric potentials at points

a

,

b

, and

c

are given as

30 \text{ V}

,

12 \text{ V}

, and

2 \text{ V}

, respectively. Determine the currents flowing through each of the three resistors.
A portion of an electric circuit is depicted in the figure. The electric potentials at points a , b , and c are given

A portion of an electric circuit is depicted in the figure. The electric potentials at points a , b , and c are given

Quantrex Academy · Accepted Answer

Correct answer: C. 1 A through 10 , 0.4 A through 20 , and 0.6 A through 30

Let the electric potential at the junction of the three resistors be

V

.

Applying Kirchhoff's Current Law (KCL) at the junction, the sum of currents leaving the junction is zero. Assuming current flows from

a

to the junction, and from the junction to

b

and

c

:

\dfrac{V_a - V}{10} = \dfrac{V - V_b}{20} + \dfrac{V - V_c}{30}

Given that

V_a = 30 \text{ V}

,

V_b = 12 \text{ V}

, and

V_c = 2 \text{ V}

, we substitute these values into the equation:

\dfrac{30 - V}{10} = \dfrac{V - 12}{20} + \dfrac{V - 2}{30}

Multiplying the entire equation by

60

to clear the denominators:

6(30 - V) = 3(V - 12) + 2(V - 2)

180 - 6V = 3V - 36 + 2V - 4

180 - 6V = 5V - 40

11V = 220 \implies V = 20 \text{ V}

Now, we calculate the current through each resistor:

Current through the

10 \text{ }\Omega

resistor:

I_{10} = \dfrac{V_a - V}{10} = \dfrac{30 - 20}{10} = 1 \text{ A}

Current through the

20 \text{ }\Omega

resistor:

I_{20} = \dfrac{V - V_b}{20} = \dfrac{20 - 12}{20} = \dfrac{8}{20} = 0.4 \text{ A}

Current through the

30 \text{ }\Omega

resistor:

I_{30} = \dfrac{V - V_c}{30} = \dfrac{20 - 2}{30} = \dfrac{18}{30} = 0.6 \text{ A}

A portion of an electric circuit is depicted in the figure. The...

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