Question

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p$ = 1$\Omega$ as shown in the figure. An external resistance of $R_e$ = 2$\Omega$ is connected via the sliding contact.

• A. 0.3 A
• B. 1.35 A
• C. 1.0 A
• D. 0.9 A

Answer 1

Answer: (C)

1.0 A

Answer 2

Solution Explanation: We model the circuit as follows:

• Let the potentiometer (total resistance 1 Ω) be split into two equal resistances: R₁ = 0.5 Ω (from battery positive to sliding contact) and R₂ = 0.5 Ω (from sliding contact to battery negative).
• Let V_A = 0.9 V (battery positive) and V_B = 0 V (battery negative).
• The external resistor Rₑ = 2 Ω is connected from the sliding contact (node C) to the negative terminal.
• Using Kirchhoff’s Current Law (KCL) at node C, where the currents coming from R₁ equal the currents leaving via R₂ and Rₑ, we write:

  (V_A − V_C) / R₁ = V_C / R₂ + V_C / Rₑ

Substitute the known values:

  (0.9 − V_C) / 0.5 = V_C / 0.5 + V_C / 2

Solve the equation step by step:

1. Multiply through by a common factor (say 2) to eliminate fractions:   2×[(0.9 − V_C) / 0.5] = 2×[V_C / 0.5] + 2×[V_C / 2]   4(0.9 − V_C) = 4V_C + V_C         (since 2/0.5 = 4 and 2/2 = 1)

2. Expand and solve:   3.6 − 4V_C = 5V_C   3.6 = 9V_C   V_C = 0.4 V

3. Now, the battery current is just the current flowing through R₁ (from the battery to node C):   I₁ = (V_A − V_C) / R₁ = (0.9 − 0.4) / 0.5 = 0.5 / 0.5 = 1.0 A

Thus, the battery supplies a current of 1.0 A.