Question

A 15.0-Ω resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed switch.
(a) At 2.00 ms after the switch is opened the current has decayed to 0.210 A. Calculate the inductance of the coil.

(b) Calculate the time constant of the circuit.

(c) How long after the switch is opened will the current reach 1.00% of its original value?

Answer 1

Answer:

0.04328 H

0.00288 seconds

0.01326 seconds

Explanation:

V = Voltage = 6.3 V

R = Resistance = 15 Ω

L = Inductance

I = Decayed current = 0.21 A

t = Time to decay = 2 ms

Maximum Current is given by

$I_0=\frac{V}{R}\\\Rightarrow I_0=\frac{6.3}{15}\\\Rightarrow I_0=0.42\ A$

Current in the circuit is given by

$I=I_0exp\frac{-Rt}{L}\\\Rightarrow ln\frac{I_0}{I}=\frac{Rt}{L}\\\Rightarrow L=\frac{Rt}{ln\frac{I_0}{I}}\\\Rightarrow L=\frac{15\times 2\times 10^{-3}}{ln\frac{0.42}{0.21}}\\\Rightarrow L=0.04328\ H$

The inductance is 0.04328 H

Time constant is given by

$\tau=\frac{L}{R}\\\Rightarrow \tau=\frac{0.04328}{15}\\\Rightarrow \tau=0.00288\ s$

Time constant is 0.00288 seconds

Time required is given by

$t=\tau ln\frac{I_0}{I}\\\Rightarrow t=0.00288\times ln\frac{I_0}{0.01I_0}\\\Rightarrow t=0.01326\ s$

The time to reach 1% of its original value is 0.01326 seconds