inductance

Question

A buld rated of 100V, 100 W it can be treated as a resistor . Find out inductance of an inductor called choke coil that should be connected in series with bulb to operate the bulb at its rated power with the help of an ac source 200V and 50Hz 

Answer 1

Hi Samagra!

 

Here is my initial take on this problem, which may change if I were to think about it more.

 

The bulb and coil are in series, under a source providing 200 V across both of them.  If the bulb is meant to have 100 V across it, then the other 100 V must be dropped across the coil.

 

The voltage across the coil is given by:

 

V_L = L(di/dt)

 

where L is the inductance, and (di/dt) is the time rate of change of the current.  Since this is AC, (di/dt) will end up depending on a sinusoidal function, but we will then (as with the other rapidly oscillating quantities under AC) convert it to a "maximum" value which will characterize it for this circuit.

 

To determine that current, we have to look at the whole circuit, and use the AC version of Ohm's law:

 

V = iZ --> i = V/Z

 

(where we usually characterize all of these as rms values to get an "average," or as maximum values)

 

with Z as the impedance of the combination.  Since this is a resistor and an inductor in series, the impedance becomes:

 

Z = √(R² + X_L²)

 

where R = resistance of the resistor (the bulb) and X_L = the inductive reactance of the coil = 2πfL, where L is again the inductance, and f is the frequency of the AC driving.

 

Now we have to calculate both R and X_L.  Let's start with R.  We know the bulb is rate to consume 100 W of power at 100 V.  Therefore, we can write:

 

P = V²/R  (P = power) -->  100 W = (100 V)²/R --> R = 100 Ω

 

Then, X_L = 2π(50 Hz)L = 100π(L)

 

So, Z = √[(100 Ω)²+ (100πL)^2]

 

Thus, if we take 200 V to be the maximum voltage amount in the circuit, we then have:

 

i_max = V_max/Z = (200 V)/√[(100 Ω)²+ (100πL)²]

 

So i(t) = i_maxsin(2πft)  (no phase shift needed here, as the current should be in phase with the driving voltage)

 

Thus, (di/dt) = i_max(2πf)cos(2πft)

 

The maximum value for (di/dt) is then:  (2πf)i_max = (100π)(200 V)/√[(100 Ω)²+ (100πL)²]=(20,000π)/√[(100²+ (100πL)²]

 

Now we can go back to the expression for voltage across an inductor, and solve for L:

 

V_L = L(di/dt) --> 100 V = L(20,000)/√[(100²+ (100πL)²] --> 100 = 20,000πL/√(10,000+10,000π²L²)

 

We can solve this expression for L.  I would start by inverting both sides, and then squaring both sides, to get:

 

(1/10,000) = 10,000+(10,000π²L²)/(4x10⁸)π²L²

 

4x10⁴π²L² = 10,000+10,000π²L²

 

From here, you can combine terms and solve for L.

 

No guarantees that I have not missed something here, but this looks to be a path to a solution.  There is quite a bit going on, so just let me know if you have any questions.  Hope this helps!