Question

An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?

Answer 1

Step-by-step explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

`X_L=0.2* 12=2.4` ohm

`X_C=(1)/(\omega * C)=(1)/(12* 0.043)=1.9379\ ohm`

`Z=\sqrt{R^2+\left ( \omega L-(1)/(\omega C) \right )^2}`

`Z=√(5^2+\left ( \2.4-1.9379 )^2)`

=5.021 ohm

so amplitude of current =
`(v)/(z)=(120)/(5.021)=23.89`