Emf of generator is 6 V and internal resistance is 0.5 k Ω. If internal resistance of voltmeter is 2.5 k Ω. then reading of voltmeter must be

Question

Emf of generator is 6 V and internal resistance is 0.5 k Ω. If internal resistance of voltmeter is 2.5 k Ω. then reading of voltmeter must be (A) 10-3 V (B) 1 V (C) 5 V (D) 10 V

Tardigrade Admin · Accepted Answer

Current flowing through the circuit I=(V/R+G) =(6/(2.5+0.5) × 1000)=(1/500) A ∴ Reading of voltmeter, V=I × G=(1/500) × 2.5 × 1000=5 volt

Q. Emf of generator is $6 V$ and internal resistance is $0.5 k Ω$ . If internal resistance of voltmeter is $2.5 k Ω$ . then reading of voltmeter must be

10-3 V

1 V

5 V

10 V

Current flowing through the circuit $I = \frac{V}{R + G}$
$= \frac{6}{( 2.5 + 0.5 ) \times 1000} = \frac{1}{500} A$
$∴$ Reading of voltmeter,
$V = I \times G = \frac{1}{500} \times 2.5 \times 1000 = 5$ volt

Q. Emf of generator is 6V and internal resistance is 0.5kΩ. If internal resistance of voltmeter is 2.5kΩ. then reading of voltmeter must be

10-3 V

1 V

5 V

10 V

Current flowing through the circuit I=R+GV​ =(2.5+0.5)×10006​=5001​A ∴ Reading of voltmeter, V=I×G=5001​×2.5×1000=5 volt

Q. Emf of generator is $6 V$ and internal resistance is $0.5 k Ω$ . If internal resistance of voltmeter is $2.5 k Ω$ . then reading of voltmeter must be

Current flowing through the circuit $I = \frac{V}{R + G}$
$= \frac{6}{( 2.5 + 0.5 ) \times 1000} = \frac{1}{500} A$
$∴$ Reading of voltmeter,
$V = I \times G = \frac{1}{500} \times 2.5 \times 1000 = 5$ volt