Question

A coil is connected in series with a 19.0 kΩ resistor. An ideal 50.0 V battery is applied across the two devices, and the current reaches a value of 2.20 mA after 2.80 ms. (a) Find the inductance of the coil.

Answer 1

inductance of the coil 29.3767 H

Step-by-step explanation:

given data

resistor R = 19.0 kΩ = 19 × 10³ Ω

potential applied V = 50.0 V

current I = 2.20 mA = 2.20 ×
`10^(-3)` A

time t = 2.80 ms = 2.80 ×
`10^(-3)` s

solution

we know for maximum current in circuit that is

current = V ÷ R .........1

current =
`(50)/(19* 10^3)`

current = 2.63 ×
`10^(-3)` A

so at time t = 0

t =
`-(L)/(R) ln(1-(I_f)/(I_(max)))`

`2.80 * 10^(-3) = -(L)/(19* 10^3) ln(1-(2.20* 10^(-3))/(2.63* 10^(-3))})`

L = 29.3767