Question

, switch S1 is closed while switch S2 is kept open. The inductance is L = 0.160 H , and the resistance is R = 120 Ω .

When the current has reached its final value, the energy stored in the inductor is 0.300 J . What is the emf E of the battery?

After the current has reached its final value, S1 is opened and S2 is closed. How much time does it take for the energy stored in the inductor to decrease to a half of the original value?

Answer 1

U = 0.300 J

we know that v = R

U = 1/2LI^2

I = sqrt(2U/L)

I = sqrt[(2 x 0.300) / 0.160 ]

I = 1.936 A

= 1.936 x 120

= 232.4 V

let the value of current is i in the circuit when the energy stored in the inductor decreases to half the original value.

U' = 1/2U

1/2LI'^2 = 1/2(1/2LI^2)

I' = I / sqrt(2)

I' = / sqrt(2)*R

we know that while discharging, I = (/R)*e^-(R/L)

/ sqrt(2)*R =  (/R)*e^-(R/L)

1 / sqrt(2) = e^-(R/L)

-(R/L)*t = ln(1/sqrt(2))

-(R/L)*t = -0.346

t = (0.346 x L) / R

t = (0.346 x 0.160) / 120

t = 461.33 us

answered by: anonymous