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

Answer:

inductance of the coil 29.3767  H

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 =  $\frac{50}{19\times 10^3}$

current = 2.63 × $10^{-3}$ A

so at time t = 0

t = $-\frac{L}{R} ln(1-\frac{I_f}{I_{max}})$  

$2.80 \times 10^{-3} = -\frac{L}{19\times 10^3} ln(1-\frac{2.20\times 10^{-3}}{2.63\times 10^{-3}}})$  

L = 29.3767