Animated Solution for Physics - Electromagnetic Induction: An arc lamp requires a direct current of $10\text{ A}$ at $80\text{ V}$ to function. If it is connected to a $220\text{ V}$ (rms), $50\text{ Hz}$ AC supply, the series inductor needed for it to work is close to

Why We Use Choke Coils

Imagine you have a powerful arc lamp that shines brilliantly, but it's a bit of a diva. It strictly demands a direct current (DC) of $10\text{ A}$ at $80\text{ V}$. If you give it more, it burns out. If you give it less, it won't light up. Now, you only have a standard household AC supply of $220\text{ V}$ at $50\text{ Hz}$. How do you safely connect this lamp without destroying it? This is a classic engineering problem that tests our understanding of AC circuits and impedance.

First, we need to understand the electrical characteristics of the arc lamp. Since it operates on DC, we can model it as a simple resistor. Using Ohm's Law ($V=IR$), we can find its resistance.

So, our arc lamp is essentially an $8\Omega$ resistor that needs exactly $10\text{ A}$ of current to function properly.

If we connect this $8\Omega$ lamp directly to the $220\text{ V}$ AC supply, the current would be $I=\frac{220}{8}=27.5\text{ A}$. This massive current would instantly destroy the lamp! We need to drop the excess voltage.

We could use a series resistor, but resistors dissipate energy as heat ($P=I^{2}R$). Dropping that much voltage across a resistor would waste a huge amount of power. Instead, we use an inductor (often called a choke coil). An ideal inductor provides opposition to alternating current (called inductive reactance) but consumes zero average power. It's the perfect, energy-efficient solution!

When we connect the inductor in series with the lamp, we create an L-R series circuit. The total opposition to current in this circuit is the impedance ($Z$). We know the circuit must allow exactly $10\text{ A}$ of current to flow from the $220\text{ V}$ supply.

The impedance of an L-R series circuit is given by the phasor addition of resistance and inductive reactance:

Substituting our known values:

Now, it's just a matter of algebra. Let's square both sides to remove the square root:

We know that inductive reactance is related to inductance ($L$) and frequency ($f$) by the formula $X_L=2\pi fL$. Plugging in the frequency of $50\text{ Hz}$:

Since $\sqrt{420}$ is approximately $20.49$, we get:

Thus, an inductor of approximately $0.065\text{ H}$ is required to safely operate the arc lamp on the AC supply.