Question

For the bridge circuit shown in figure, the voltages are:

$V_1=\sqrt{2}cos1000tV,$

$V_2=2cos(1000t+{45}^{\circ })and{V}_d=0.$

The value of $R=100\Omega .\text{Then}{Z}_x\text{will be}$

• A. $100\Omega$ resistor in series with 100 mH inductor
• B. $100\Omega$ resistor in series with $100\mu F$ capacitor
• C. $100\Omega$ resistance in parallel with 100 mH inductor
• D. $100\Omega$ resistance in parallel with $100\mu F$ capacitor.

Answer 1

The correct option is A $100\Omega$ resistor in series with 100 mH inductor



$V_1=\sqrt{2}cos1000t$

$V_2=2cos(1000t+{45}^{\circ })$

$V_d=0$

$Here,\omega =1000$

Now in loop 1, writing KVL

$V_1-IR+V_d=0$

$V_1=IR$

$I=\frac{V_1}{R}=\frac{\sqrt{2}}{100}cos1000t$ ... (i)

Now in loop 2,

$V_2-V_d-I{Z}_x=0$

$2cos(1000t+{45}^{\circ })=\frac{\sqrt{2}}{100}\left(cos1000t\right)\times Z_x$

So both will be equal if $Z_x$ provide 45° phase and magnitude $100\sqrt{2}.$

$i.e.Z_x=100+100j$

So a resistance of $100\Omega$ in series with inductor providing reactance $=100\Omega$

$\omega L=100\Omega$

$L=\frac{100}{\omega }=0.1H=100mH$