Question 19.SP.1: A series circuit contains a resistance of 20 Ω and an induct......
A series circuit contains a resistance of 20 Ω and an inductance of 10 H connected across a voltage source of 110 V. (a) What is the current 1 s after the circuit is closed? (b) What are v_{R} and v_{C} at this time?
Some time after the current reaches a steady value, the switch is opened. When the series circuit is open, the RL circuit opposes the decay of current toward the steady-state value of zero. (c) What is the current 2 s after the circuit is opened? (d) What are v_{R} and v_{C} at this time?
(a) Step 1. Write the formula for charging or rising current when the switch is closed.
i=\frac{V}{R}\left(1-e^{-R t / L}\right) (19-2)
Step 2. Find the value of e^{-R t / L} for t = 1 s.
-\frac{R t}{L}=-\frac{20(1)}{10}=-2 \quad e^{-2}=0.135Step 3. Substitute values for e^{-R t / L}, V, and R in Eq. (19-2).
i=\frac{110}{20}(1-0.135)=5.5(0.865)=4.76\ A(b) Write the formulas for v_R and v_{ L } after the switch is closed and solve them when t = 1 s.
v_R=V\left(1-e^{-R t / L}\right) (19-5)
= 110(0.865) = 95.2 V
Also v_{R} = iR = 4.76(20) = 95.2 V.
v_{ L }=V e^{-R t / L}= 110(0.135) = 14.8 V
Check: V=v_R+v_L (19-4)
110 = 95.2 + 14.8
110 V = 110 V
(c) The steady-state value of current is
I=\frac{V}{R}=\frac{110}{20}=5.5\ AWrite the formula for discharging or decaying current when the switch is opened and substitute values with t = 2 s.
i=\frac{V}{R} e^{-R t / L} (19-7)
\begin{aligned}-\frac{R t}{L} & =-\frac{20(2)}{10}=-4 \\ e^{-4} & =0.018 \end{aligned}i = 5.5(0.018) = 0.10 A
(d) Write the formulas for v_{R} and v_{L} after the switch is opened and solve them when t = 2 s.
v_R=V e^{-Rt / L} (19-10)
= 110(0.018) = 1.98 V
v_L=-V e^{-R t / L} (19-11)
= -110(0.018) = -1.98 V
Check: 0=v_R+v_L (19-9)
0 = 1.98 – 1.98
0 V = 0 V