Problem #$1$

Circle the correct answer.

$1.$ for tsec$,$ 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

$2.$ for tsec$,$ 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$=3$sec equals:

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

$($b$)2$ J

$($c$)4$ J

$($d$)8$ J

$($e$)20$ J

$3.$ 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

$4.$ 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

$5.$ Given v$($t$)=5\backslash$sqrt$(2)$cos$(40$t$)$V$,$ the average value of this signal is:

$($a$)0$ V

$($b$)5$ V

$($c$)5\backslash$sqrt$(2)$V

$($d$)8\backslash$sqrt$(2)$V

$($e$)20$ V

$6.$ Given v$($t$)=5\backslash$sqrt$(2)$cos$(40$t$)$V$,$ the r$.$m$.$s value of this signal is:

$($a$)0$ V

$($b$)5$ V

$($c$)5\backslash$sqrt$(2)$V

$($d$)8\backslash$sqrt$(2)$V

$($e$)20$ V

$7.$ If the voltage across load x is given by v$\_($x$)($t$)=8$sin$(4$t$+30^(0))$V$,$ while the current through

this load is i$\_($x$)($t$)=2$sin$(4$t$-60\backslash$deg $)$ A$.$ then this load is a:

$($a$)$ resistor

$($b$)$ capacitor

$($c$)$ inductor

$($d$)$ resistor and capacitor

$($e$)$ resistor and inductor