In an LR circuit the coil has an of 30ml and the resistor has a resistance of 40

Question

In an LR circuit the coil has an of 30ml and the resistor has a resistance of 40 ohm At t = 0 they are to a 24V battery (a) How long does it take for the current to of its final value? (b) what is the power in the resister at this maximum? A proton, which is initially at rest, is released from the positive plate of a capacitor and acquires a velocity of v = 10 degree m/s when it hits the negative plate, If the capacitance is C = lnF how much charge is residing on the capacitor plates? The shadow of a 3m high stick, standing vertically, is 3m long on the ground, if the same stick is placed, also standing vertically, in a flat-bottomed tank filled with 2m of some liquid, the shadow is 2.25m long on the bottom of the tank. What is the index of refraction of the liquid? Can you identify it?

Explanation / Answer

Resistance R = 40 ohm

Inductance L = 30 mH = 0.03 H

Voltage V = 24 volt

Maximum current = io

In LR circuit , current i = io[1- e-Rt/L]

According to problem , i = (2/3) io

Substitute values you get ,(2/3) io = io [1- e-Rt/L]

  [1- e-Rt/L] = 2/3

[e-Rt/L] = 1-(2/3)

= 1/3

-Rt/L = ln(1/3)

= -1.098

t = 1.098 L / R

= 1.098(0.03)/40

= 8.239 x10 -4 s

(b). Power dissipated P = i 2 R

Where i = (2/3)io = (2/3) (V/R)

= (2/3)(24/40)

= 0.4 A

So, P = (0.4) 2 (40)

= 6.4 watt

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