Question

A battery of emf \( 8V \) and internal resistance \( 0.5 \Omega \) is being charged by a \( 120V \) DC supply using a series resistor of \( 15.5 \Omega \). The terminal voltage of the \( 8V \) battery during charging is:

• A. \( 11.5V \)
• B. \( 1.15V \)
• C. \( 115V \)
• D. \( 0.5V \)

Hint:

During charging, the terminal voltage of a battery is higher than its emf due to the voltage drop across the internal resistance.

Answer 1

The Correct Option is A

Step 1: Using Ohm’s Law for the Charging Circuit
The total resistance in the circuit is:
\[ R_{\text{total}} = R_{\text{series}} + R_{\text{internal}} = 15.5 + 0.5 = 16 \Omega \] The charging current \( I \) is given by: \[ I = \frac{V_{\text{supplied}} - V_{\text{battery}}}{R_{\text{total}}} \] \[ I = \frac{120 - 8}{16} = \frac{112}{16} = 7A \] Step 2: Calculating the Terminal Voltage
The terminal voltage \( V_{\text{terminal}} \) of the battery during charging is:
\[ V_{\text{terminal}} = V_{\text{battery}} + I R_{\text{internal}} \] \[ V_{\text{terminal}} = 8 + (7 \times 0.5) = 8 + 3.5 = 11.5V \] Step 3: Conclusion
Thus, the terminal voltage of the \( 8V \) battery during charging is \( \mathbf{11.5V} \).