Question

For the circuit shown below,

the voltage across the capacitor $\left(V_c\right)$ is

• A. $1.05\angle {24}^{\circ }V$
• B. $10.15\angle {24}^{\circ }V$
• C. $15.10\angle {24}^{\circ }V$
• D. $5.10\angle {24}^{\circ }V$

Answer 1

The correct option is B $10.15\angle {24}^{\circ }V$

Using de-coupled technique,

Applying KVL in loop-1,
$10\angle 0^{\circ }=I_1(5+j5-j10)+I_2(j2-j10)$
$=(5-j5)L_1+(-j8)I_2$ ...(i)
Applying KVL in loop-2,
$10\angle {90}^{\circ }=I_2(5+j5-j10)+I_1(j2-j10)$
$10\angle {90}^{\circ }=I_2(5-j5)+(-j8)I_1$ ...(ii)
Now adding equation (i) and (ii),
$10+j10=I_1(5-j13)+I_2(5-j13)$
$I_1+I_2=\frac{10+j10}{5-j13}$
$V_c=-j{X}_C(I_1+I_2)$
$=-j10\times \frac{10+j10}{5-j13}=10.15\angle {24}^{\circ }V$