Four different resistors are arranged as shown. A current of 1.5 A enters the network at junction X and leaves through junction Y.

Question 19

Four different resistors are arranged as shown.

A current of 1.5 A enters the network at junction X and leaves through junction Y.

What is the current in the resistor of resistance 30 Ω?

A 0.21 A                      B 0.50 A                     C 0.75 A                     D 1.0 A

Reference: Past Exam Paper – June 2018 Paper 12 Q34

Solution:

Answer: B.

The resistors may be re-arranged as follows

Let the combined resistance be R.

1 / R = 1/70 + 1/(50+30+60)

Combined resistance R = 46.67 Ω

The p.d. across the parallel connection is

V = IR

V = 1.5 × 46.67 = 70 V

From Ohm’s law: I = V / R

Current in the 70 Ω resistor, I₁ = 70 / 70 = 1.0 A

At junction X, the current is split and flows in each branch.

I₁ + I₂ = 1.5

So, the current in the lower branch,

I₂ = 1.5 – 1.0 = 0.50 A