Consider an L-R circuit as shown in the figure. (Figure 1)The battery provides 1
ID: 1342590 • Letter: C
Question
Consider an L-R circuit as shown in the figure. (Figure 1)The battery provides 12.0 V of voltage. The inductor has inductance L, and the resistor has resistance R = 150 ?. The switch is initially open as shown. At time t=0, the switch is closed. At time t after t=0 the current I(t)flows through the circuit as indicated in the figure.
(a) What is the voltage reading Vr(t) given by the voltmeter across the resistor (Figure 2) at time tafter t=0?
(b) What is the voltage reading Vind(t) given by the voltmeter across the inductor (Figure 3) at time t after t=0? In the following expressions, ?I/?t denotes the time rate of change of the current in the inductor at time t.
(c) After the switch is closed, the current in the circuit grows over time approaching a constant value. In general, at time t after a voltage source is connected to an L-R circuit, the current I(t) in the circuit is given by the expression
I(t)=VR(1?e?t/?),
where V is the voltage provided by the battery, R is the resistance of the resistor, and ? is the time constant characteristic of the circuit.
What is the current reading I(?) given by the ammeter shown in the circuit (Figure 4) at one time constant after t=0?
Explanation / Answer
Current I(t) = ( V/R ) ( 1 - e -t/)
Current at t =0 is : I(0) = (V/R) (1 - e -0) = (V/R) (1 -1 ) = 0
(a) Vr(t = 0) = I (0) x R = 0 Amp
(b) Vind (t = 0) = [ Vbattery - Vr (0) ] = [ 12 - 0 ] = 12 Volt
(c) At one time constant t = T
I(t = T ) = (V/R)( 1 - e-T/T ) = (V/R) (1 - e-1) = (V/R) (1 - 1/e) = (12/150)( 1 - 1/2.718) = 0.05 Amp = 50 mili Amp