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

Answer:

14.488 amperes

Step-by-step explanation:

The amplitude I of the current is given by

$\large I=\displaystyle\frac{E_m}{Z}$

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

$\large Z=\sqrt{R^2+(X_L-X_C)^2}$

where

R= Resistance

L = Inductance

C = Capacitance

w = Angular frequency

$\large X_L=wL$ = inductive reactance

$\large X_C=\displaystyle\frac{1}{wC}$ = 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

$\large Z=\sqrt{7^2+(4.44-0.012)^2}=8.283$

Finally

$\large I=\displaystyle\frac{E_m}{Z}=\frac{120}{8.283}=14.488\;amperes$