An alternating current E(t)=120sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance of L=0.37 henrys, a resistor of R=7ohms and a cap…

Question

asked Sep 7, 2020 151k views

1 Answer

Paul Hunnisett · Answer 1 · 2020-09-11T18:39:26+0000

Answer:

14.488 amperes

Explanation:

The amplitude I of the current is given by

$\large I=\displaystyle(E_m)/(Z)$

where

$\large E_m$ = amplitude of the energy source E(t).

Z = Total impedance.

The amplitude of the energy source is 120, the maximum value of E(t)

The total impedance is given by

$\large Z=√(R^2+(X_L-X_C)^2)$

where

R= Resistance

L = Inductance

C = Capacitance

w = Angular frequency

$\large X_L=wL$ = inductive reactance

$\large X_C=\displaystyle(1)/(wC)$ = capacitive reactance

As E(t) = 120sin(12t), the angular frequency w=12

So

$\large X_L=12*0.37=4.44\\\\X_C=1/(12*7)=0.012$

and

$\large Z=√(7^2+(4.44-0.012)^2)=8.283$

Finally

$\large I=\displaystyle(E_m)/(Z)=(120)/(8.283)=14.488\;amperes$