Problem #$2$

$(5$ points$)$

For the circuit in figure $(1).$ Given $i_s(t)=8cos(2t+{30}^0)A,R=4,C=0\mathrm{.}25F,L=2H,$ Circle the correct answer.

The total impedance seen by the source is

$($a$)(1+j2)$

$($b$)(1-j2)$

$($c$)1$

$($d$)2$

$($e$)8$

The operating frequency of the given circuit is

$($a$)1$ Hz

$($b$)2$ Hz

$($c$)Hz$

$($d$)2Hz$

$($e$){}^{-1}Hz$

The amplitude $($magnitude$)$ of the capacitor voltage $v_c(t)$ is:

$($a$)2$ V

$($b$)4$ V

$($c$)8$ V

$($d$)16$ V

$($e$)32$ V

The phase of the capacitor voltage $v_c(t)$ is:

$($a$)-{60}^0$

$($b$)30$

$($c$)-90$

$($d$)90$

$($e$){120}^0$

Replace the capacitor with a short circuit. So$,$ the current source, inductor and resistor are connected together in parallel. Let the current through the inductor $i_L(t)$ to be the output. The equation relating the input and the output is given by $d{i}_L\frac{t}{d}t+i_L(t)=i_s(t).$ Then $$ equals:

$($a$)0\mathrm{.}125$

$($b$)0\mathrm{.}5$

$($c$)2$

$($d$)4$

$($e$)8$