Animated Solution for Physics - Current Electricity: An electric bulb of $500\text{ W}$ at $100\text{ V}$ is used in a circuit having a $200\text{ V}$ supply. Calculate the resistance $R$ to be connected in series with the bulb, so that the power delivered by the bulb is $500\text{ W}$.

A Bulb in Danger

Imagine you have just bought a brilliant, high-intensity electric bulb. The packaging proudly states that it delivers $500\text{ W}$ of power when connected to a $100\text{ V}$ source. It is a beautiful piece of engineering.

But there is a catch. The only power supply you have available in your laboratory is a $200\text{ V}$ source.

What happens if you connect this $100\text{ V}$ bulb directly to a $200\text{ V}$ supply? The physics is unforgiving. The bulb's internal filament is designed to handle a specific maximum current. Doubling the voltage would force double the current through the filament, quadrupling the power dissipated as heat (since $P=\frac{V^2}{R}$). In a fraction of a second, the filament would vaporize in a bright flash. Boom.

To save the bulb, we need a "bodyguard"—a component that can absorb the excess voltage before it reaches the bulb. This is where a series resistor comes into play. Our mission is to calculate the exact resistance $R$ required to keep the bulb operating safely at its rated $500\text{ W}$.

Understanding the Bulb's Appetite

Before we can protect the bulb, we need to understand its electrical characteristics. The bulb is rated for $500\text{ W}$ at $100\text{ V}$.

Power in an electrical circuit is the product of voltage and current. The fundamental equation is:

We can rearrange this equation to find the exact current the bulb needs to operate optimally:

Substituting the rated values of our bulb:

This tells us that for the bulb to shine perfectly without burning out, exactly $5\text{ A}$ of current must flow through it. Not a milliampere more, not a milliampere less.

The Series Circuit Strategy

We know we must connect a resistor $R$ in series with the bulb. Why in series?

In a series circuit, the current has only one path to flow. This means the current flowing through the resistor will be exactly the same as the current flowing through the bulb. Therefore, the current through our protective resistor will also be $5\text{ A}$.

Furthermore, in a series circuit, the total supply voltage is divided among the components. According to Kirchhoff's Voltage Law (KVL), the sum of the voltage drops across the components must equal the total supply voltage.

Mathematically, this is expressed as:

We know the supply voltage is $200\text{ V}$, and we want the bulb to receive exactly its rated $100\text{ V}$.

Let's find out how much voltage the resistor needs to absorb:

Our bodyguard resistor must drop exactly $100\text{ V}$ across its terminals.

Calculating the Resistance

Now we have all the pieces of the puzzle for our resistor. We know the voltage it must drop ($100\text{ V}$) and the current that will flow through it ($5\text{ A}$).

To find the resistance, we turn to the most famous law in electronics: Ohm's Law. Ohm's Law states that the voltage across a resistor is directly proportional to the current flowing through it, with the resistance being the constant of proportionality:

We can rearrange this to solve for the unknown resistance $R$:

Substituting our known values:

And there is our answer! To safely operate the $100\text{ V}$ bulb on a $200\text{ V}$ supply, we must connect a $20\Omega$ resistor in series.

Energy Efficiency

While we have successfully solved the problem and saved the bulb, it is worth pausing to consider the engineering implications of this solution.

We added a resistor to drop $100\text{ V}$ at $5\text{ A}$. Let's calculate the power dissipated by this resistor:

The resistor is dissipating $500\text{ W}$ of power as pure heat! This means our total circuit is consuming $1000\text{ W}$ of power from the supply, but only $500\text{ W}$ is being used by the bulb to produce light. The other $500\text{ W}$ is completely wasted as heat by the resistor.

In a real-world scenario, this is highly inefficient. A $500\text{ W}$ resistor would be massive, expensive, and require significant cooling to prevent it from catching fire. While this series resistor method works perfectly on paper and in textbook problems, modern electronics would use much more efficient methods, such as a step-down transformer (for AC circuits) or a switching DC-DC converter, to achieve the same result without the massive energy waste.

But for the sake of mastering circuit analysis and Ohm's Law, this classic problem remains an elegant and essential exercise!