In the given circuit the internal resistance of the 18 V cell is negligible. If $\mathrm{R}_{1}=400 \Omega, \mathrm{R}_{3}=100 \Omega$ and $\mathrm{R}_{4}= 500 \Omega$ and the reading of an ideal voltmeter across $\mathrm{R}_{4}$ is 5 V , then the value of $\mathrm{R}_{2}$ will be

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Current Electricity

In the given circuit the internal resistance of the 18 V cell is negligible. If

R_{1} = 400Ω, R_{3} = 100Ω

and

R_{4} = 500Ω

and the reading of an ideal voltmeter across

R_{4}

is 5 V , then the value of

R_{2}

will be

Answer: (A)

Work backwards from the voltmeter. It reads

5 V

across

R_{4} = 500 Ω

, so the current there is

i_{4} = \frac{5}{500} = 0.01 A

, and since

R_{3}

is in series with

R_{4}

the same current drops

V_{3} = 0.01 \times 100 = 1 V

across it. The

R_{3}

R_{4}

branch therefore sits at

6 V

, leaving

18 - 6 = 12 V

across

R_{1}

, so the current in

R_{1}

i_{1} = \frac{12}{400} = 0.03 A

. By the junction rule the current turning into the

R_{2}

branch is

i_{2} = 0.03 - 0.01 = 0.02 A

against a

6 V

drop, giving

R_{2} = \frac{6}{0.02} = 300 Ω

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option