A battery of emf 8 V with internal resistance `0.5Omega` is being charged by a 120 V dc supply using a series resistance of `15.5Omega`. The terminal voltage of the battery is

Options

A. 20.5 V
B. 15.5 V
C. 11.5 V
D. 2.5 V

Explanation

To find the terminal voltage of the battery while it is being charged, we can follow these steps:

### Step 1: Identify the given values
- EMF of the battery (E) = 8 V
- Internal resistance of the battery (r) = 0.5 Ω
- External DC supply voltage (V_s) = 120 V
- Series resistance (R) = 15.5 Ω

### Step 2: Calculate the effective voltage
The effective voltage (V_eff) that is used to charge the battery can be calculated by subtracting the EMF of the battery from the external supply voltage:
\[ V_{eff} = V_s - E = 120 \, \text{V} - 8 \, \text{V} = 112 \, \text{V} \]

### Step 3: Calculate the total resistance in the circuit
The total resistance (R_total) in the circuit is the sum of the series resistance and the internal resistance of the battery:
\[ R_{total} = R + r = 15.5 \, \Omega + 0.5 \, \Omega = 16 \, \Omega \]

### Step 4: Calculate the current in the circuit
Using Ohm's law, we can find the current (I) flowing through the circuit:
\[ I = \frac{V_{eff}}{R_{total}} = \frac{112 \, \text{V}}{16 \, \Omega} = 7 \, \text{A} \]

### Step 5: Calculate the terminal voltage of the battery
The terminal voltage (V_t) of the battery while charging can be calculated using the formula:
\[ V_t = I \cdot r + E \]
Substituting the values we found:
\[ V_t = (7 \, \text{A} \cdot 0.5 \, \Omega) + 8 \, \text{V} = 3.5 \, \text{V} + 8 \, \text{V} = 11.5 \, \text{V} \]

### Final Answer
The terminal voltage of the battery while being charged is **11.5 V**.

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