time elapsed in an LR circuit

Question

An LR circuit having time constant of 50 ms is connected with an ideal battery of emf E . prove that the time elapsed before the power dissipated in heat reaches half its maximum value is 61ms

Answer 1

Hi Samagra!

 

The power dissipated will (ideally) all be in the resistor.  Because there is a standard expression for the rise of current in a DC series LR circuit (which I assume this is, if it has a time constant):

 

i(t) = (V/R)(1 - e^-t/τ_L)

 

where

 

V = battery emf

τ_L = time constant

 

we can write the expression for power dissipated in the resistor using I and R.

 

P = I²R

 

At long times, the above expression for current shows that i = V/R.  This would be the same as if there were only a resistor in the circuit.  This is because, after long times, the current is no longer changing much, and the voltage across the inductor depends on a changing current.  If the current is stable, the inductor is just a coiled piece of wire (assumed ideal, with no resistance, in our case).

 

The maximum power dissipated in the resistor will thus be when the current is maximum, meaning after long times, when I = V/R.  At that point, we have:

 

P = V²/R

 

The target power we need to calculate the time for is half of this value:  P_1/2 = V²/2R

 

When the power has this value, what is the current?

 

P_1/2 = V²/2R = (I_1/2)²R

I_1/2² = V²/2R²

I_1/2 = (1/√2)(V/R)

 

So, now, we want to solve for the time during "power up" at which the current equals this value.

 

(1/√2)(V/R) = (V/R)(1 - e^-t/50ms)

1/√2 = 1 - e^-t/50ms

e^-t/50ms = 1 - 1/√2 = 0.293

-t/0.050 s = ln(0.293) = -1.23

t = (1.23)(0.050) = 0.0615 s = 61.5 ms

 

I hope this helps! Just let me know if you have any questions.