NCERT Solutions

We need to calculate the terminal voltage of a charging battery and explain the role of the series resistor.

Concept: Terminal voltage

The terminal voltage ($V$) of a battery during charging is higher than its electromotive force ($E$) due to the voltage drop across its internal resistance ($r$). The circuit follows Ohm's Law and Kirchhoff's Voltage Law.

• Charging current:

$$I = \frac{V_{\text{supply}} - E}{R + r}$$

• Terminal voltage:

$$V = E + Ir$$

Given:

• EMF of the battery, $E = 8.0\,\text{V}$
• Internal resistance of the battery, $r = 0.5\,\Omega$
• DC supply voltage, $V_{\text{supply}} = 120\,\text{V}$
• External series resistance, $R = 15.5\,\Omega$

Solving:

1. Calculate the net EMF in the circuit:

   Since the battery is being charged, the supply voltage opposes the battery's EMF.

   $$\begin{aligned} E_{\text{net}} &= V_{\text{supply}} - E \\ &= 120\,\text{V} - 8.0\,\text{V} \\ &= 112.0\,\text{V} \end{aligned}$$

2. Calculate the total resistance of the circuit:

   The series resistor and internal resistance are in series.

   $$\begin{aligned} R_{\text{total}} &= R + r \\ &= 15.5\,\Omega + 0.5\,\Omega \\ &= 16.0\,\Omega \end{aligned}$$

3. Calculate the charging current ($I$):

   Using Ohm's law for the entire circuit:

   $$\begin{aligned} I &= \frac{E_{\text{net}}}{R_{\text{total}}} \\ &= \frac{112.0\,\text{V}}{16.0\,\Omega} \\ &= 7.0\,\text{A} \end{aligned}$$

4. Calculate the terminal voltage ($V$) of the battery:

   During charging, the terminal voltage is the sum of the EMF and the potential drop across the internal resistance.

   $$\begin{aligned} V &= E + Ir \\ &= 8.0\,\text{V} + (7.0\,\text{A} \times 0.5\,\Omega) \\ &= 8.0\,\text{V} + 3.5\,\text{V} \\ &= 11.5\,\text{V} \end{aligned}$$

Result:

The terminal voltage of the battery during charging is

$$11.5\,\text{V}$$

The purpose of the series resistor is to limit the charging current. Without it, the low internal resistance of the battery would result in an extremely high current from the $120\,\text{V}$ supply, which could overheat and damage the battery.