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
. Using Hook's law,
, 