In balanced meter bridge, the resistance of bridge wire is 0.1 Ω/ cm. Unknown resistance X is connected in the left gap and 6 Ω in right gap. Null point divides the wire in the ratio 2: 3. The current drawn from the battery of 5 V having negligible resistance is (a) 1 A (b) 1.6 A (c) 2 A (d) 5 A

Answer: (a). X/6 = 2/3 ⇒ X = 4 Ω; gap branch 4 + 6 = 10 Ω Wire: 100 × 0.1 = 10 Ω Req = 10 ∥ 10 = 5 Ω I = 5/5 = 1 A So the answer is (a) 1 A.

In balanced meter bridge, the resistance of bridge wire is 0.1 Ω/ cm… | MHT CET 2026 (14 May, Shift 2)

Q9MHT CET 2026 · 14 MAY · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

In balanced meter bridge, the resistance of bridge wire is $0.1\ \Omega/$ cm. Unknown resistance $X$ is connected in the left gap and $6\ \Omega$ in right gap. Null point divides the wire in the ratio $2 : 3$ . The current drawn from the battery of $5$ V having negligible resistance is

Show answer and solution

Answer: (a)

$\dfrac{X}{6} = \dfrac{2}{3} \Rightarrow X = 4\ \Omega$ ; gap branch $4 + 6 = 10\ \Omega$
Wire: $100 \times 0.1 = 10\ \Omega$
$R_{eq} = 10 \parallel 10 = 5\ \Omega$
$I = \dfrac{5}{5} = 1$ A
So the answer is (a) $1$ A.

Report a mistake

Wheatstone Bridge & Metre Bridge question from MHT CET 2026 (14 May, Shift 2)

More questions on Wheatstone Bridge & Metre Bridge