Question

A series RL circuit is built with 190 Ω resistor and a 5.0-cm-long, 1.0-cm-diameter solenoid with 800 turns of wire.

A) What is the peak magnetic flux through the solenoid if the circuit is driven by a 12 V, 5.0 kHz source?

B) What capacitance, in μF, has its potential difference increasing at 1.5×10⁶ V/s when the displacement current in the capacitor is 0.60 A ?

Answer 1

What is the peak magnetic flux through the solenoid if the circuit is driven by a 12 V, 5.0 kHz source

first , we find the inductance , L

L = u_o N² A / L

where

A = pi * 0.5e-2² = 7.853e-5 m²

so,

L = 4e-7 * 800² * 7.853e-5 / 0.05

L = 1.2633e-3 H

Now,

I = V / sqrt ( R² + (2fL)² )

I = 12 / sqrt ( 190² + (2 * 5000 * 1.2633e-3)² )

I = 0.0618 A

so,

peak flux

= IL

= 7.81e-5 Wb (PLEASE MIND THE UNITS !!!)

_____________________________

What capacitance, in μF, has its potential difference increasing at 1.5×10⁶ V/s when the displacement current in the capacitor is 0.60 A

dV / dt = 1.5e6 V/s

we know

dQ / dt = C * dV / dt = I

so,

I = C * dV / dt

C = I / (dV / dt)

C = 0.6 / 1.5e6

C = 4e-7 F

OR

C = 0.4 uF (PLEASE MIND THE UNITS !!!)