Question

In the circuit shown, if $v\left(t\right)=2sin\left(1000t\right)$ volts. $R=1k\Omega$ and $C=1\mu F$. then the steady-state current $i\left(t\right)$, in milliamperes (mA), is

• A. $3sin\left(1000t\right)+cos\left(1000t\right)$
• B. $sin\left(1000t\right)+cos\left(1000t\right)$
• C. $sin\left(1000t\right)+3cos\left(1000t\right)$
• D. $2sin\left(1000t\right)+2cos\left(1000t\right)$

Answer 1

The correct option is A $3sin\left(1000t\right)+cos\left(1000t\right)$


Here, $X_C=\frac{1}{\omega C}=\frac{1}{{10}^3\times {10}^{-6}}=\frac{1}{{10}^{-3}}$

$X_C={10}^3\Omega$

$R={10}^3\Omega$ (Given)

$v\left(t\right)=2sin1000tV=2\angle 0^{\circ }V$

Redrawing the given network, we get,

As the bridge is balanced, it can be redrawn as

$\therefore Y_{eq}=Y_1+Y_2$

$=\frac{3}{2R}+\frac{1}{-2j{X}_C}$

$=\frac{3}{2}\times {10}^{-3}+j\frac{1}{2}\times {10}^{-3}$

$\therefore i\left(t\right)=v\left(t\right)\times Y_{eq}=2\angle 0^{\circ }[\frac{3}{2}+j\frac{1}{2}]mA$

$=(3+j1)mA$

$=3sin\left(1000t\right)+cos\left(1000t\right)mA$