Question

A battery of emf 10 V and internal resistance 3 W is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer 1

given, emf of battery, E = 10 V

internal resistance of battery, r = 3Ω

battery is connected to a resistor of resistance R (let).

from Kirchoff's law,

E - ir - iR = 0

or, E = i(r + R)

given, current through circuit , i = 0.5A

so, 10V = 0.5A(3Ω + R)

or, 20 = 3Ω + R

or, R = 17Ω

so, resistance of the resistor, R = 17Ω

Terminal voltage , V = E - ir

= 10 V - 0.5A × 3Ω

= 10 V - 1.5 V

= 8.5 V

hence terminal voltage of the battery when the circuit is closed = 8.5V

Answer 2

AnswEr:-

• The resistance of the resistor is 17 Ω.
• Terminal voltage of the resistor is 8.5 V.

Explanation:-

Emf of the battery, E = 10 V

Internal resistance of the battery, r= 3Ω

Current in the circuit, I = 0.5 A

Resistance of the resistor= R

The relation for current using Ohm's Law

→ I = R/R+r

→ R+r= E/I

→ 10/9.5

→ 20 Ω

•°• R = 20-3 = 17Ω

Now, Terminal voltage of the resistor= V

A/Q Ohm's Law

→ V= IR

→ 0.5×17

→ 8.5 V

$\huge\ddot{\smile}$