Figure shows a part of an electric circuit. The potentials at points $\mathrm{a}$, $\mathrm{b}$ and c are $30 \mathrm{~V}$, $12 \mathrm{~V}$ and 2 V respectively. The current through the $20 \Omega$ resistor will be.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Current Electricity

Figure shows a part of an electric circuit. The potentials at points

a

b

and c are

30 V

12 V

and 2 V respectively. The current through the

20Ω

resistor will be.

Answer: (A)

Let the potential at the common junction be

V

. The three resistors meet there, so Kirchhoff's current law gives

\frac{V - 30}{10} + \frac{V - 12}{20} + \frac{V - 2}{30} = 0

. Clearing denominators,

6 (V - 30) + 3 (V - 12) + 2 (V - 2) = 0

, which gives

11 V = 220

, so

V = 20

V. The current in the

20 Ω

resistor is

\frac{V - 12}{20} = \frac{20 - 12}{20} = 0.4

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option