3. In an LR circuit the coil has an inductance of 30 mH, the resistor has an unk

Question

3. In an LR circuit the coil has an inductance of 30 mH, the resistor has an unknown resistance. They are connected in series to a 9 V battery at t=0. (a) It takes 69 mu s for the energy stored in the solenoid to reach 25% of its final value (Attention: energy not current. Calculate . (b) At what rate is energy being dissipated in the resistor at this moment? (c) At what rate the magnetic energy stored ?n the coil is increasing at this moment? 4. (a) What is the average output power of the Sun if the peak magnetic field of sunlight is 3.4x 10^5T on Earth? (b) How long does it take for sunlight to deliver 25kWh energy (same as a 25W light bulb radiates in 1000 hours) to a 1 cm^2 area on the surface of the Earth? (Assume the mean distance of Earth from Sun is 150 million kilometers.)

Explanation / Answer

3)

L = 30 mH

For,current,

I = Io(1 - e^(-t/T))

where , Io = final current after a long time = V/R = 9/R

T = time constant = L/R

Energy stored, U = 0.5*L*I^2

So, I = sqrt(2U/L)

So, I/Io = sqrt(U/Uo)

For, energy to be 25 percent( or 0.25 times) of final value Uo,

So, I/Io = sqrt(0.25*Io/Io) = sqrt(0.25) = 0.5 times

So, I = 0.5*Io

So, 0.5*Io = Io*(1-e^(-t/T))

So, t/T = 0.693

So, t = 0.693*T = 0.693*L/R

So, 69*10^-6 = 0.693*40*10^-3/R

So, R = 401.7 ohm <------answer

b)

Rate of dissipation of energy = I^2*R = (Io*(1-e^(-t/T)))^2*R

= ((9/401.7)*(1-e^(-69*10^-6/(30*10^-3/401.7))))^2*401.7

= 0.0733 W

c)

rate of change of magnetic energy = 0.5*2*L*I*(dI/dt) = L*I*dI/dt

= L*I*(Io*(t/T)*e^(-t/T))

= 30*10^-3*(0.0135)*(9/401.7)*(69*10^-6/(30^10^-3/401.7))*e^(-(69*10^-6/(30^10^-3/401.7)))

= 2.4*10^-7 W

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