Question

In the circuit shown, reading of the voltmeter connected across $400\Omega$ resistance is $30\text{V}.$ If it is connected across $300\Omega$ resistance then reading will be $\frac{900}{x}\text{V}.$ The value of $x$ is (integer only)

Answer 1

Given, potential drop across $V_{400\Omega }=30\text{V}$

case - I


$\Rightarrow \text{potential drop across}{V}_{300\Omega }=30\text{V}$

$\Rightarrow I=\frac{30}{300}=\frac{1}{10}\text{A}$

$\Rightarrow I_1=\frac{30}{400}=\frac{3}{40}\text{A}$

Using applying $\text{KVL}$

$30=(I-I_1)R_V$

$\Rightarrow R_V=1200\Omega$

Case - II



Net current flow through the circuit is,

$I=\frac{60}{400+\frac{300\times 1200}{1200+300}}=\frac{3}{32}\text{A}$

Using applying $\text{KVL}$

$300I_0=(I-I_0)1200$

$\Rightarrow I_0=\frac{4I}{5}\text{A}$

$\Rightarrow I_0=\frac{4}{5}\times \frac{3}{32}=\frac{3}{40}\text{A}$

The reading of voltmeter is,

$V=\frac{3}{40}\times 300=\frac{900}{40}$

But the given reading is $V=\frac{900}{x}$

$\therefore x=40$