Question

The reading of the voltmeter (rms)in volts, for the circuit shown in the figure is

• A. 141.42

Answer 1

The correct option is A 141.42
Given bridge is shown in the figure:

Form the given bridge circuit, we see that product of opposite arm impedances are equal i.e.$(-j)\times (-j)=\left(j\right)\left(j\right)$

$or,-1=-1$

Hence, the bridge is balanced, i.e. no current flows through the voltage meter.

Now,

$V_i\left(t\right)=100sin\omega t$

$=\frac{100}{\sqrt{2}}Volt\left(rmsvalue\right)$

$\therefore$ Net current supplied by the source is
$I=\frac{100/\sqrt{2}}{(0.5+Z_{eq})}A$

$Z_{eq}=0(i.e.netimpedanceofwholebridge=0)$

$\therefore I=\frac{100/\sqrt{2}}{0.5}=141.42A$

$\therefore \frac{I}{2}=70.71A$

Current flows through each parallel path of the bridge circuit therefore,

$V=V_a-V_b$

$=(\frac{I}{2}\times -j)-(\frac{I}{2}\times j)$

$=-jI=-j141.42volt\left(rmsvalue\right)$

$\therefore$ Reading of the voltmeter in volt (rms)

$=V=V_{ab}=141.42V$