A solenoid with internal resistance is connected in series with a 20-? resistor

Question

A solenoid with internal resistance is connected in series with a 20-? resistor and a battery of negligible internal resistance and a switch. The above graph shows a voltage measurement as a function time after the switch is closed. (a) Across which circuit element is the voltage being measured? Explain. (You must explain to earn a full credit for this question.) (b) What is the internal resistance of this solenoid? (c) What is the maximum current in this circuit? (d) What is the inductance of this solenoid?

Explanation / Answer

a)

The voltage being measured is across solenoid. Because at t=0 all of the voltage is acroos the solenoid.

b)

The graph shoes that the voltage across the solenoid reches finaly to 10 V, therefore:

isolenoid = iresistor

10/Rs = 50/(R+Rs)

10/Rs = 50/(20+Rs)

>>>> Rs = 5 ohm

c)

i = V/(R+Rs) = 50/(20+5) = 2 A

d)

V = 40 (e^-Rt/L) + 10

at t=2ms >>> V = 25 V

>>>> 25 = 40*(e^(-25*2e-3/L)) + 10

>>>> L = 0.0510 H

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