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

Explanation of the Solution:

1. Divide the potentiometer: The potentiometer of $R_p = 1\,\Omega$ is tapped at the midpoint so that each half has a resistance of $0.5\,\Omega$.

2. Assign Node Voltages: Let the battery voltage be $V = 0.9\,V$ with terminal A at $0.9\,V$ and terminal B at $0\,V$. Let node C be the tap point.

3. Currents Through Resistors:

   • Current through the top half: $$I_1 = \frac{0.9 - V_C}{0.5}$$
   • Currents from node C: $$I_2 = \frac{V_C}{0.5} \quad \text{(through the lower half)}$$ $$I_3 = \frac{V_C}{2} \quad \text{(through external resistor $ R_e = 2\,\Omega $)}$$

4. Apply Kirchhoff’s Current Law (KCL) at Node C:

   $$I_1 = I_2 + I_3$$

   Substituting the expressions:

   $$\frac{0.9 - V_C}{0.5} = \frac{V_C}{0.5} + \frac{V_C}{2}$$

5. Solve for $V_C$: Multiply the whole equation by 2 (LCM of 0.5 and 2):

   $$2\times\frac{0.9 - V_C}{0.5} = 2\times\frac{V_C}{0.5} + 2\times\frac{V_C}{2}$$

   Since $2/0.5 = 4$ and $2/2 = 1$, we get:

   $$4(0.9 - V_C) = 4V_C + V_C$$ $$3.6 - 4V_C = 5V_C \quad \Longrightarrow \quad 3.6 = 9V_C \quad \Longrightarrow \quad V_C = 0.4\,V$$

6. Determine $I_1$ (Total current from the battery):

   $$I_1 = \frac{0.9 - 0.4}{0.5} = \frac{0.5}{0.5} = 1.0\,A$$