A battery of EMF 10 V, with internal resistance `1 Omega` is being charged by a 120 V d.c. supply using a series resistance of ` 10 Omega`. The terminal voltage of the battery is

Options

A. 20 V
B. 10 V
C. Zero
D. 30 V

Answer

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

### Step 1: Identify the components and their values
- EMF of the battery (E) = 10 V
- Internal resistance of the battery (r) = 1 Ω
- External charging supply voltage (V_supply) = 120 V
- Series resistance (R) = 10 Ω

### Step 2: Set up the equation using Kirchhoff's Voltage Law (KVL)
When charging the battery, we can apply KVL around the loop:
\[ V_{supply} - E - I \cdot r - I \cdot R = 0 \]
Where:
- \( V_{supply} \) is the external voltage supply (120 V)
- \( E \) is the EMF of the battery (10 V)
- \( I \) is the current flowing through the circuit
- \( r \) is the internal resistance of the battery (1 Ω)
- \( R \) is the external resistance (10 Ω)

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ 120 - 10 - I \cdot 1 - I \cdot 10 = 0 \]
This simplifies to:
\[ 110 - 11I = 0 \]

### Step 4: Solve for the current (I)
Now, we can solve for the current \( I \):
\[ 11I = 110 \implies I = \frac{110}{11} = 10 \text{ A} \]

### Step 5: Calculate the terminal voltage (V)
The terminal voltage \( V \) of the battery while charging can be calculated using the formula:
\[ V = E + I \cdot r \]
Substituting the known values:
\[ V = 10 + 10 \cdot 1 = 10 + 10 = 20 \text{ V} \]

### Conclusion
The terminal voltage of the battery when being charged is **20 V**.

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