Question

A battery of emf \(15\ \text{V}\) and internal resistance of \(4\ \Omega\) is connected to a resistor. If the current in the circuit is \(2\ \text{A}\), the resistance of the resistor and terminal voltage of the battery will be:

• A. \(2.5\ \Omega,\ 6\ \text{V}\)
• B. \(3.5\ \Omega,\ 6\ \text{V}\)
• C. \(2.5\ \Omega,\ 7\ \text{V}\)
• D. \(3.5\ \Omega,\ 7\ \text{V}\)

Hint:

Remember: \[ E=I(R+r) \] and \[ V=E-Ir \]

• Terminal voltage decreases when current flows
• Internal resistance consumes part of emf

Answer 1

The Correct Option is B

Concept: For a cell with internal resistance: \[ E=I(R+r) \] where: \[ E = \text{emf of battery} \] \[ R = \text{external resistance} \] \[ r = \text{internal resistance} \] Terminal voltage is: \[ V = IR \]

Step 1: Calculate external resistance. Given: \[ E=15\ \text{V} \] \[ r=4\ \Omega \] \[ I=2\ \text{A} \] Using: \[ E=I(R+r) \] \[ 15=2(R+4) \] \[ 15=2R+8 \] \[ 2R=7 \] \[ R=3.5\ \Omega \]

Step 2: Calculate terminal voltage. \[ V=IR \] \[ V=2\times3.5 \] \[ V=7\ \text{V} \] Therefore, the correct answer is: \[ \boxed{3.5\ \Omega,\ 7\ \text{V}} \]