A battery is connected in series with a 0.29-? resistor and an inductor, as show

Question

A battery is connected in series with a 0.29-? resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is 0.18 s, and the maximum current in the circuit is8.3 A.

(a) Find the emf of the battery.
V

(b) Find the inductance of the circuit.
mH

(c) Find the current in the circuit after one time constant has elapsed.
A

(d) Find the voltage across the resistor after one time constant has elapsed.
V

(e) Find the voltage across the inductor after one time constant has elapsed.
V

A battery is connected in series with a 0.29-? resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is 0.18 s, and the maximum current in the circuit is8.3 A. a) Find the emf of the battery. V (b) Find the inductance of the circuit. mH (c) Find the current in the circuit after one time constant has elapsed. A (d) Find the voltage across the resistor after one time constant has elapsed. V (e) Find the voltage across the inductor after one time constant has elapsed. V

Explanation / Answer

Voltage = V

Inductance = L =

Resistance = R = 0.29 ohm

Maximum Current = V/R = 8.3 A

Time constant = L/R = 0.18

a) EMF of the battery = R*(V/R) = 0.29*8.3 = 2.407 V

b) Inductance = R*(L/R) = 0.29*0.18 = 0.0522 H = 52.2 mH

c) Current in the circuit after 1 time constant = (V/R)*(1 - exp(-t/T) = 8.3*(1-exp(-1)) = 5.246 A

d) Voltage across resistor after 1 time constant = V*(1 - exp(-t/T) = V*(1 - exp(-T/T)

= 2.407*(1-exp(-1)) = 1.522 V

e) Voltage across inductor after 1 time constant = V*exp(-t/T) = 2.407*exp(-T/T)

= 2.407*exp(-1) = 0.885 V

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