Let Îµ0= 8 V, R = 3.0 , and L = 10.5 H. The switch is closed attime t = 0.

At time t = 0.5 s, find the rate at which the battery suppliespower.

A picture of the circuit can be found here:
http://tinypic.com/view.php?pic=ekq3ih&s=7

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Hubert KochLv2

28 Feb 2019

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Consider an L-R circuit as shown in the figure.Thebattery provides 12.0 V of voltage. The inductor hasinductanceL, and the resistor has resistanceR= 150 Î©. The switch is initiallyopen as shown. Attime t = 0, the switch is closed. At time t aftert = 0 the currentI(t) flows through the circuit as indicated in thefigure.

Part A:

What is the voltagechange V_r(t)across the resistor at time tafter t = 0?

What is the voltage changeV_ind(t)across the inductor at time t after t = 0? In thefollowingexpressions, ÎI/Ît denotes the time rate of change ofthecurrent in the inductor at time t?

After the switch is closed, thecurrentin the circuit grows over time approaching a constant value.Ingeneral, at time t after a voltage source is connected to anL-Rcircuit, the current I(t) in the circuit is given bytheexpression

I(t)=V/R(1-e^-t/Ï)

where V is the voltage provided bythebattery, R is the resistance of the resistor, and Ï is thetimeconstant characteristic of the circuit.

Part C:

What is the current I(Ï) flowing inthecircuit shown in the figure at one time constant after t =0?

Express your answer in amperes.

Hint: To find the current inthecircuit after one time constant, use the expression forI(t)given in the problem, with t equal to Ï. Do not try todeterminethe time constant for this circuit. You would not be ableto do sobecause the inductance L is unknown. Instead, afterplugging in thenumerical values for V and R appropriatefor this circuit andsubstituting Ï for time, simplifyyour expression to get a numericalvalue for the current.

In the following figure(Figure 1) switch, S1 is closed while switch S2 is kept open. The inductance is L = 0.380 H, the resistance is R = 61.0 Ω, and the emf of the battery is 18.0 V. At time t after S1 is closed, the current in the circuit is increasing at a rate of di / dt = 7.20 A/s.