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Question

Physics Question on Current electricity

A DC source of 30V30 \, V is connected across two resistors of 200Ω200\, \Omega each connected in series. A voltmeter of resistance 200Ω200 \, \Omega is connected across a resistor. It reads

A

20 V

B

30 V

C

10 V

D

5 V

Answer

10 V

Explanation

Solution

Given situation is shown in the figure
Net resistance of the circuit
R=200+200×200200+200R = 200 + \frac{200 \times 200}{200 + 200}
=300Ω= 300 \Omega
Current drawn from battery,
I=VR=30300=0.1AI= \frac{V}{R} = \frac{30}{300} = 0.1 A
I1=200×I200+200=I2,I_1 = \frac{200 \times I}{200+200} =\frac{I}{2}, or, I2=l2I_2 = \frac{l}{2}
Reading of voltmeter =I1R=(0.12)×200=10V = I_1 R = \left(\frac{0.1}{2}\right) \times 200 = 10 \, V