A portion of a circuit is illustrated in the figure. If there is a current of $12 \text{ mA}$ in the $5 \text{ k}\Omega$ resistor, determine the currents in the remaining three resistors and the potential difference between points $A$ and $B$ .

Question

A portion of a circuit is illustrated in the figure. If there is a current of

12 \text{ mA}

in the

5 \text{ k}\Omega

resistor, determine the currents in the remaining three resistors and the potential difference between points

A

and

B

.

Quantrex Academy · Accepted Answer

Correct answer: A. 4 mA in 20 k resistor, 8 mA in 10 k resistor, 12 mA in 100 k resistor; 1340 V

The total current entering the circuit is

I = 12 \text{ mA}

, which flows through the

5 \text{ k}\Omega

resistor.

This current splits into the parallel combination of

20 \text{ k}\Omega

and

10 \text{ k}\Omega

resistors. Using the current divider rule:
Current in the

20 \text{ k}\Omega

resistor is

I_{20} = I \times \dfrac{10}{20 + 10} = 12 \times \dfrac{10}{30} = 4 \text{ mA}

.
Current in the

10 \text{ k}\Omega

resistor is

I_{10} = I \times \dfrac{20}{20 + 10} = 12 \times \dfrac{20}{30} = 8 \text{ mA}

.

The currents recombine, so the total current of

12 \text{ mA}

flows through the

100 \text{ k}\Omega

resistor.

The equivalent resistance of the circuit between A and B is:

R_{eq} = 5 + \left(\dfrac{20 \times 10}{20 + 10}\right) + 100 = 105 + \dfrac{20}{3} = \dfrac{335}{3} \text{ k}\Omega

The potential difference between points A and B is:

V_{AB} = I \times R_{eq} = 12 \text{ mA} \times \dfrac{335}{3} \text{ k}\Omega = 4 \times 335 = 1340 \text{ V}

Answer:

4 \text{ mA}

in

20 \text{ k}\Omega

resistor,

8 \text{ mA}

in

10 \text{ k}\Omega

resistor,

12 \text{ mA}

in

100 \text{ k}\Omega

resistor;

1340 \text{ V}

A portion of a circuit is illustrated in the figure. If there is a...