Question

A 8 $\Omega$ resistor is connected to a battery that has an internal resistance of 0.2 $\Omega$. If the voltage across the battery (the terminal voltage) is 10 V, then the emf of the battery is

• A. 10.15 V
• B. 10.20 V
• C. 10.25 V
• D. 9.80 V

Hint:

Terminal voltage of a discharging battery: $V_T = E - Ir$.
Emf of a battery: $E = V_T + Ir$.
Current in the circuit: $I = \frac{E}{R+r} = \frac{V_T}{R}$.
The terminal voltage $V_T$ is the voltage across the external resistor $R$.

Answer 1

The Correct Option is C

Let $E$ be the emf of the battery, $r$ be its internal resistance, and $R$ be the external resistance. The terminal voltage ($V_T$) across the battery when it is discharging (supplying current to $R$) is given by: $V_T = E - Ir$, where $I$ is the current flowing in the circuit. The current $I$ flowing through the external resistor $R$ is also related to the terminal voltage by Ohm's law: $V_T = IR$. So, $I = \frac{V_T}{R}$. Given values: External resistance $R = 8 \text{ } \Omega$. Internal resistance $r = 0.2 \text{ } \Omega$. Terminal voltage $V_T = 10 \text{ V}$. First, calculate the current $I$: $I = \frac{V_T}{R} = \frac{10 \text{ V}}{8 \text{ } \Omega} = \frac{10}{8} \text{ A} = \frac{5}{4} \text{ A} = 1.25 \text{ A}$. Now, use the terminal voltage equation to find the emf $E$: $V_T = E - Ir \Rightarrow E = V_T + Ir$. $E = 10 \text{ V} + (1.25 \text{ A}) \times (0.2 \text{ } \Omega)$. $Ir = 1.25 \times 0.2 = 1.25 \times \frac{2}{10} = \frac{2.50}{10} = 0.25 \text{ V}$. So, $E = 10 \text{ V} + 0.25 \text{ V} = 10.25 \text{ V}$. \[ \boxed{10.25 \text{ V}} \]