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:
A real battery consists of an ideal electromotive force (EMF, \(E\)) and an internal resistance (\(r\)). The terminal voltage and current vary depending on the external load connected to it.
Step 2: Key Formula or Approach:
Ohm's law for a complete circuit with internal resistance is:
\[ E = I(R + r) \] We can write this equation for the two different scenarios and solve the resulting system of equations to find \(r\).
Step 3: Detailed Explanation:
Scenario 1: \(I_1 = 0.6 \text{ A}\), \(R_1 = 3 \, \Omega\)
\[ E = 0.6(3 + r) \] Scenario 2: \(I_2 = 0.4 \text{ A}\), \(R_2 = 6 \, \Omega\)
\[ E = 0.4(6 + r) \] Since the battery's EMF (\(E\)) is the same in both cases, equate the two expressions:
\[ 0.6(3 + r) = 0.4(6 + r) \] Divide both sides by 0.2 to simplify:
\[ 3(3 + r) = 2(6 + r) \] Expand the brackets:
\[ 9 + 3r = 12 + 2r \] Isolate \(r\):
\[ 3r - 2r = 12 - 9 \] \[ r = 3 \, \Omega \] Step 4: Final Answer:
The internal resistance of the battery is 3 \(\Omega\).