Question

The equivalent capacitance of the input loop of the circuit shown is

• A. $2\mu F$
• B. $100\mu F$
• C. $200\mu F$
• D. $4\mu F$

Answer 1

The correct option is A $2\mu F$


Applying KVL,

$V_{in}-i_1(1+1)-50i_1(-j{X}_c)=0$

$\Rightarrow V_{in}=i_1[2-j50X_c]$

$Inputimpedance=\frac{V_{in}}{i_1}=2-j50X_c$

As imaginary part is negative, input impedance has equivalent capacitive reactance $X_{Ceq}$

$X_{Ceq}=50X_c$

$\frac{1}{\omega C_{eq}}=\frac{50}{\omega C}=\frac{50}{\omega \times 100}=\frac{1}{2\omega }$

$C_{eq}=2\mu F$