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:
The potentiometer (total resistance 1 Ω) is tapped at its midpoint so that the resistances from the tap to the ends are 0.5 Ω each. The battery (0.9 V) is connected across the entire 1 Ω, and an external resistor $R_e = 2\,\Omega$ is connected from the tap (which has potential $V_{tap}$) to the negative terminal.

1. The current through the resistor from the positive terminal to the tap is:

$$I_{R1} = \frac{0.9 - V_{tap}}{0.5}$$

2. The currents leaving the tap are:
   • Through the lower half of the potentiometer: $$I_{R2} = \frac{V_{tap}}{0.5}$$
   • Through the external resistor: $$I_{e} = \frac{V_{tap}}{2}$$

Apply KCL at the tap:

$$\frac{0.9 - V_{tap}}{0.5} = \frac{V_{tap}}{0.5} + \frac{V_{tap}}{2}$$

Multiply through by 2 (since $\frac{2}{0.5} = 4$):

$$4(0.9 - V_{tap}) = 4V_{tap} + V_{tap}$$ $$3.6 - 4V_{tap} = 5V_{tap}$$ $$3.6 = 9V_{tap}$$ $$V_{tap} = \frac{3.6}{9} = 0.4\,V$$

Now, the battery current is:

$$I = I_{R1} = \frac{0.9 - 0.4}{0.5} = \frac{0.5}{0.5} = 1\,A$$