Multiple Choice
An LR circuit with L = 0.1 H and R = 10 Ω are connected to a battery with the circuit initially broken. When the circuit is closed, how much time passes until the current reaches half of its maximum value?
30. Induction and Inductance
LR Circuits
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Consider the LR circuit shown below. Initially, both switches are open. Switch 1 is closed. a) What is the maximum current in the circuit after a long time? Then, S1 is opened and S2 is closed. b) What is the current in the circuit after 0.05s?
A
(a) imax=0.5 A;
(b) i (0.05s)=1.86×10−6 A
B
(a) imax=2 A;
(b) i (0.05s)=7.45×10−6 A
C
(a) imax=0.5 A;
(b) i (0.05s)=0.14 A
D
(a) imax=2 A;
(b) i (0.05s)=0.57 A
E
(a) imax=0.5 A;
(b) i (0.05s)=0.46 A A
F
(a) imax=2 A;
(b) i (0.05s)=1.8 A
Verified step by step guidance1
To find the maximum current in the circuit after a long time when switch S1 is closed, we need to consider the circuit as a simple series circuit with a resistor and a battery. The inductor will act as a short circuit after a long time.
Use Ohm's Law to find the maximum current: I_max = V / R, where V is the voltage of the battery (10 V) and R is the resistance (5 Ω).
Now, when switch S1 is opened and S2 is closed, the circuit becomes an RL circuit with the inductor and resistor in series. The current will decay over time according to the formula: I(t) = I_max * e^(-Rt/L), where I_max is the initial current, R is the resistance, L is the inductance, and t is the time.
Substitute the known values into the decay formula: I(0.05s) = I_max * e^(-5 * 0.05 / 0.02).
Calculate the exponent and then the current at 0.05 seconds using the exponential decay formula.
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