A battery is connected in series with a 0.31- resistor and an inductor, as shown

Question

A battery is connected in series with a 0.31- resistor and an inductor, as shown in the figure below. The switch is closed at t = o. The time constant of the circuit is 0.12 s, and the maximum current in the circuit is 7.8 A. (a) Find the emf of the battery. (b) Find the inductance of the circuit. mH (c) Find the current in the circuit after one time constant has elapsed. (Round your answer to four decimal places.) (d) Find the voltage across the resistor after one time constant has elapsed. (Round your answer to four decimal places.) (e) Find the voltage across the inductor after one time constant has elapsed.

Explanation / Answer

Time constant of RL circuit

T = L/R = 0.12 sec

=> L = 0.12*0.31 = 37.2 mH OPTION (b)

Max. current in the circuit = 7.8 A

at t = very long time

inductor will act as a short circuit

=> I(max) = E/R = 7.8 A

=> emf of the battery E = 7.8*0.31 = 2.418 volt OPTION(a)

Equation of charging RL circuit

I(t) = Io*(1 - e^(-t/T))

after t = T

=> I(T) = Io*(1 - e^(-1)) = 0.632*Io = 0.632*7.8 =4.929 A OPTION(c)

Voltage across Resistor after t = T

V = I(T)*R = 4.929*0.31 = 1.527 volt OPTION (d)

Voltage across inductor after t = T

V(T) = E*e^(-t/T) = E*0.367 = 2.418*0.367 = 0.887 volt OPTION(e)

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