Five resistors of resistance $1\Omega ,1\Omega ,2\Omega ,3\Omega$ and $4\Omega ,$ a battery of voltage 10V, an ideal ammeter and five switches $S_1,S_2,S_3,S_4$ and $S_5$ are connected in a circuit as shown in the figure. When $S_3$ and $S_5$ are closed and $S_1,S_2$ and $S_4$ remain opened then current read by ammeter is $i_1.$ When $S_1,S_3$ and $S_5$ are closed and $S_2,S_4$ remain opened then current read by ammeter is $i_2.$ The ratio of current $\frac{i_1}{i_2}$ is given by $0.11\times \lambda .$ Find the value of $\lambda .$

Solution

Case 1 – S3 and S5 closed

• S3 short-circuits 4Ω resistor → B = L.
• Between L and T: two 1Ω resistors in parallel → 0.5Ω.
• Path A: 0.5Ω + 2Ω = 2.5Ω
• Path B: 3Ω
• Equivalent resistance: 2.5Ω ∥ 3Ω = 15/11 Ω ≈ 1.36Ω
• Current: i₁ = 10 / (15/11) = 22/3 A ≈ 7.33 A

Case 2 – S1, S3 and S5 closed

• S1 short-circuits 1Ω, so T = L.
• S3 short-circuits 4Ω, so B = L.
• Now only 2Ω and 3Ω remain between L and R in parallel.
• Equivalent resistance: 2Ω ∥ 3Ω = 6/5 Ω = 1.2Ω
• Current: i₂ = 10 / (6/5) = 25/3 A ≈ 8.33 A

Ratio

i₁ / i₂ = (22/3) / (25/3) = 22/25 = 0.88

Given i₁/i₂ = 0.11 × λ → λ = 8

Final Answer

λ = 8