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Question

Physics Question on Electric charges and fields

Emf of generator is 6V6 \,V and internal resistance is 0.5kΩ0.5\, k \Omega. If internal resistance of voltmeter is 2.5kΩ2.5\, k \Omega. then reading of voltmeter must be

A

10-3 V

B

1 V

C

5 V

D

10 V

Answer

5 V

Explanation

Solution

Current flowing through the circuit I=VR+GI=\frac{V}{R+G} =6(2.5+0.5)×1000=1500A=\frac{6}{(2.5+0.5) \times 1000}=\frac{1}{500} A \therefore Reading of voltmeter, V=I×G=1500×2.5×1000=5V=I \times G=\frac{1}{500} \times 2.5 \times 1000=5 volt