Question

A 0.200 H inductor is connected in series with a 88.0 Ω resistor and an ac source. The voltage across the inductor is vL=−(12.0V)sin[(487rad/s)t].
Required:
a. Derive an expression for the voltage vR across the resistor. Express your answer in terms of the variables L, R, VL (amplitude of the voltage across the inductor), ω, and t .
b. What is vR at 2.00 ms? Express your answer with the appropriate units.

Answer 1

Answer:

a.  (VL)R/ωL[1 - cos[ωt]]  = (10.84 V)[1 - cos[(487rad/s)t]]

b. 1.084 mV

Explanation:

a. Since it is a series circuit, the current in the inductor is the same as the current in the resistor.

Now, the voltage across the inductor vL = -Ldi/dt.

So, the current, i = -1/L∫vLdt.

Now, vL = −(12.0V)sin[(487rad/s)t] and L = 0.200 H

Substituting these into i, we have

i = -1/L∫vLdt

= -1/0.200H∫[−(12.0V)sin[(487rad/s)t]]dt.

= -[−(12.0V)]/0.200H∫[sin[(487rad/s)t]]dt.

= 60V/H∫[sin[(487rad/s)t]]dt

Integrating i, we have

i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + C

at t = 0, i(0) = 0

0 = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)× 0]] + C

0 = 60V/H ÷ [(487rad/s)[-cos[0]] + C

0 = 60V/H ÷ [(487rad/s)[-1]+ C

C = 60V/H ÷ [(487rad/s)

So, i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + 60V/H ÷ [(487rad/s)

i =  60V/H ÷ [(487rad/s)[1 - cos[(487rad/s)t]]

i = (0.123A)[1 - cos[(487rad/s)t]] = VL/ωL[1 - cos[ωt]] where ω = 487rad/s and VL = 12.0 V and L = 0.200 H

So, the voltage across the resistor vR = iR where R = resistance of resistor = 88.0 Ω

So, vR = iR = VL/ωL[1 - cos[ωt]] × R = (VL)R/ωL[1 - cos[ωt]]

=  (0.123A)[1 - cos[(487rad/s)t]] × 88.0 Ω

= (10.84 V)[1 - cos[(487rad/s)t]]

b. vR at t = 2.00 ms = 0.002 s

So, vR = (10.84 V)[1 - cos[(487rad/s)(0.002)]]

= (10.84 V)[1 - cos[0.974]]

= (10.84 V)[1 - 0.9999]

= (10.84 V)(0.0001)

= 0.001084

= 1.084 mV