Question

A storage battery of emf \(10V\) and internal resistance \(1 \Omega\) is being charged by a \(100V\) dc supply using a series resistor of \(17 \Omega\). - The terminal voltage of the battery during charging is

• A. \(25V\)
• B. \(30V\)
• C. \(20V\)
• D. \(15V\)

Hint:

When a battery is charging, the terminal voltage is given by: \[ V_T = E + I r \] where \(I = \frac{V - E}{R + r}\). This is different from discharging mode, where \(V_T = E - I r\).

Answer 1

The Correct Option is D

Step 1: Understanding the Circuit The given circuit consists of:
- A DC supply voltage \(V = 100V\)
- A battery with emf \(E = 10V\)
- Internal resistance of the battery \(r = 1 \Omega\)
- An external series resistor \(R = 17 \Omega\) Using Ohm's Law, the current flowing through the circuit is given by: \[ I = \frac{V - E}{R + r} \] Substituting the values: \[ I = \frac{100V - 10V}{17\Omega + 1\Omega} = \frac{90}{18} = 5A \] Step 2: Finding Terminal Voltage The terminal voltage \(V_T\) of the battery during charging is given by: \[ V_T = E + I r \] Substituting the known values: \[ V_T = 10V + (5A \times 1\Omega) = 10V + 5V = 15V \] Thus, the correct answer is \( \mathbf{(4)} \ 15V \).