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
[tex]I_0=\frac{V}{R}\\\Rightarrow I_0=\frac{6.3}{15}\\\Rightarrow I_0=0.42\ A[/tex]
Current in the circuit is given by
[tex]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[/tex]
The inductance is 0.04328 H
Time constant is given by
[tex]\tau=\frac{L}{R}\\\Rightarrow \tau=\frac{0.04328}{15}\\\Rightarrow \tau=0.00288\ s[/tex]
Time constant is 0.00288 seconds
Time required is given by
[tex]t=\tau ln\frac{I_0}{I}\\\Rightarrow t=0.00288\times ln\frac{I_0}{0.01I_0}\\\Rightarrow t=0.01326\ s[/tex]
The time to reach 1% of its original value is 0.01326 seconds
