Problem #$1$

$(7$ points$)$

Circle the correct answer.

for $0t5sec,$ if the current through a $0\mathrm{.}02$ H inductor is $i_L(t)=5(t+1)A,$ then the voltage across it given by:

$($a$)0\mathrm{.}1$ V

$($b$)1$ V

$($c$)0\mathrm{.}1$ tV

$($d$)0\mathrm{.}1$ t V

$($e$)5(t+1)V$

for $0t5sec,$ if the current through a $0\mathrm{.}02$ H inductor is $i_L(t)=5(t+1)A,$ then the energy stored in the inductor at $t=3sec$ equals:

$($a$)0\mathrm{.}2$ J

$($b$)2$ J

$($c$)4$ J

$($d$)8$ J

$($e$)20$ J

Four $8$ mH inductors are connected in series, then their total inductance is:

$($a$)2$ mH

$($b$)4$ mH

$($c$)8$ mH

$($d$)16$ mH

$($e$)32$ mH

Four $8$ mF capacitors are connected in parallel, then their total capacitance is:

$($a$)2$ mF

$($b$)4$ mF

$($c$)8$ mF

$($d$)16$ mF

$($e$)32$ mF

Given $v(t)=5\sqrt[\phantom{2}]{2}cos(40t)V,$ the average value of this signal is:

$($a$)0$ V

$($b$)5$ V

$($c$)5\sqrt[\phantom{2}]{2}V$

$($d$)8\sqrt[\phantom{2}]{2}V$

$($e$)20$ V

Given $v(t)=5\sqrt[\phantom{2}]{2}cos(40t)V,$ the r$.$m$.$s value of this signal is:

$($a$)0$ V

$($b$)5$ V

$($c$)5\sqrt[\phantom{2}]{2}V$

$($d$)8\sqrt[\phantom{2}]{2}V$

$($e$)20$ V

If the voltage across load $x$ is given by $v_x(t)=8sin(4t+{30}^0)V,$ while the current through this load is $i_x(t)=2sin(4t-60)$ A$.$ then this load is a:

$($a$)$ resistor

$($b$)$ capacitor

$($c$)$ inductor

$($d$)$ resistor and capacitor

$($e$)$ resistor and inductor