Q. A DC source of $30 \, V$ is connected across two resistors of $200\, \Omega$ each connected in series. A voltmeter of resistance $200 \, \Omega$ is connected across a resistor. It reads
Solution:
Given situation is shown in the figure
Net resistance of the circuit
$R = 200 + \frac{200 \times 200}{200 + 200}$
$ = 300 \Omega$
Current drawn from battery,
$I= \frac{V}{R} = \frac{30}{300} = 0.1 A$
$I_1 = \frac{200 \times I}{200+200} =\frac{I}{2},$ or, $I_2 = \frac{l}{2}$
Reading of voltmeter $ = I_1 R = \left(\frac{0.1}{2}\right) \times 200 = 10 \, V$
