Question

1. A 2.40 mH inductor is connected in series with a dc battery of negligible internal resistance, a 0.880 kΩ resistor, and an open switch. a. How long after the switch is closed will it take for the current in the circuit to reach half of its maximum value ? b. How long after the switch is closed will it take for the energy stored in the inductor to reach half of its maximum value?

2. A 4.75 μF capacitor is initially charged to a potential of 15.4 V . It is then connected in series with a 3.60 mH inductor. a. What is the total energy stored in this circuit? b. What is the maximum current in the inductor? c. What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

3. You have a special lightbulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.50 A , even for an instant. What is the largest root-mean-square current you can run through this bulb?

4. An electric motor is being powered with a voltage amplitude of280 V at 60 Hz . The motor draws a current amplitude of 13.0 A a. Find the root-mean-square voltage. b. Find the root-mean-square current. c. Find the average power consumed by the motor.

5. A capacitance C and an inductance L are operated at the same angular frequency. a. At what angular frequency will they have the same reactance? b. If L = 4.70 mH and C = 3.30 μF , what is the numerical value of the angular frequency in part A? c. What is the reactance of each element?

Answer 1

1)

Time constant, T = L/R = 0.0024/880 = 2.73 x 10^-6 sec

A)

Since, I = I_max/2

I = I_max(1 - e^-t/T)

t = - T ln(0.5) = - (2.73 x 10^-6) ln(0.5) = 1.89 x 10^-6 sec

B)

1 - e^-t/T = 1/sqrt(2)

t = - T ln(0.293) = - 2.73 x 10^-6 x ln(0.293) = 3.35 x 10^-6 sec

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