Question

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 capacitor of capcitance C=0.037 farads.What is the amplitude of the current I?

Answer 1

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`