Question

A battery supplies $0.6 \text{ A}$ current when a $3 \Omega$ resistor is connected with it. When the resistor $3 \Omega$ is replaced by $6 \Omega$, the current is reduced to $0.4 \text{ A}$. Then, the internal resistance of the battery is

• A. $3 \Omega$
• B. $9 \Omega$
• C. $6 \Omega$
• D. $12 \Omega$

Hint:

For problems with two cases involving the same cell, the constant is EMF ($E$). Setting up the ratio of currents \( \frac{I_1}{I_2} = \frac{R_2 + r}{R_1 + r} \) is the fastest way to solve.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Every real battery has an internal resistance ($r$). When connected to an external circuit ($R$), the total resistance is $R + r$. The EMF ($E$) remains constant for the battery.
Key Formula or Approach:
Ohm's Law for a complete circuit: \( E = I(R + r) \).

Step 2: Detailed Explanation:
Let $E$ be the EMF and $r$ be the internal resistance.
Case 1: $I_1 = 0.6 \text{ A}$, $R_1 = 3 \Omega$
\[ E = 0.6(3 + r) \quad \dots \text{(Eqn 1)} \]
Case 2: $I_2 = 0.4 \text{ A}$, $R_2 = 6 \Omega$
\[ E = 0.4(6 + r) \quad \dots \text{(Eqn 2)} \]
Since $E$ is the same in both cases, equate (1) and (2):
\[ 0.6(3 + r) = 0.4(6 + r) \]
Divide by 0.2:
\[ 3(3 + r) = 2(6 + r) \]
\[ 9 + 3r = 12 + 2r \]
\[ 3r - 2r = 12 - 9 \]
\[ r = 3 \Omega \]

Step 3: Final Answer:
The internal resistance of the battery is $3 \Omega$.