An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance

Q: Why can't we just connect the bulb directly to the 200 V supply?

If you connect a 100 V rated bulb to a 200 V supply, it will draw too much current, dissipate four times its rated power, and immediately burn out!

Q: Does the resistance of the bulb change with temperature?

In reality, yes. The filament's resistance increases as it heats up. However, in standard textbook problems like this, we assume the resistance remains constant at its operating temperature.

Q: Why do we use $P = V^2/R$ instead of $P = I^2 R$ to find the bulb's resistance?

We are given the rated voltage and rated power. We don't know the current yet. So, $P = V^2/R$ is the most direct formula to find the intrinsic resistance of the bulb.

Q: What happens to the power dissipated by the resistor $R$?

The resistor $R$ also has 100 V across it and 5 A flowing through it, so it dissipates 500 W of power as heat. This is why using a series resistor to drop voltage is very inefficient in real life!

Q: Can we use a transformer instead of a resistor?

Yes! In AC circuits, a step-down transformer is a much more efficient way to reduce 200 V to 100 V without wasting 500 W of power as heat.

Question

An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance $R$ that must be put in series with the bulb, so that the bulb delivers 500 W is ......$\Omega$.

Sigma Percentile · Accepted Answer

### The Power of the Bulb

Imagine you are holding a standard electric bulb. Printed on its glass are two crucial numbers: **500 W** and **100 V**.

What do these numbers actually mean? They tell us the bulb's "comfort zone." If you provide it exactly $100	ext{ V}$, it will shine brightly, consuming exactly $500	ext{ W}$ of electrical power.

But beneath these numbers lies a hidden, intrinsic property of the bulb: its **resistance**.

Using the power formula $P = \frac{V^2}{R}$, we can uncover this hidden value.

$$R_b = \frac{V^2}{P} = \frac{(100)^2}{500} = 20\ \Omega$$

This $20\ \Omega$ resistance is a physical property of the filament. It won't change (assuming constant temperature), regardless of the circuit we put it in.

### The Circuit Dilemma

Now, we face a practical problem. Our power supply is **$200	ext{ V}$**.

If we connect our $100	ext{ V}$ bulb directly to this $200	ext{ V}$ supply, it will draw excessive current and burn out in a spectacular flash!

To protect the bulb and ensure it operates exactly at its rated $500	ext{ W}$, we must "absorb" the extra voltage. This is where the series resistor $R$ comes to the rescue.

In a series circuit, the total voltage is shared among the components.

$$V_{	ext{supply}} = V_R + V_{	ext{bulb}}$$

Since the supply is $200	ext{ V}$ and the bulb needs exactly $100	ext{ V}$, the resistor $R$ must drop the remaining voltage.

$$V_R = 200	ext{ V} - 100	ext{ V} = 100	ext{ V}$$

### The Elegant Conclusion

Take a step back and look at the circuit now.

The bulb has $100	ext{ V}$ across it. The resistor $R$ also has $100	ext{ V}$ across it.

Because they are connected in series, the **exact same current** flows through both of them.

According to Ohm's Law ($V = IR$), if two components in series have the same voltage drop, their resistances must be perfectly identical!

$$R = R_b = 20\ \Omega$$

Alternatively, we could have calculated the current first. The bulb draws $I = \frac{P}{V} = \frac{500}{100} = 5	ext{ A}$.

Applying Ohm's Law to the resistor:

$$R = \frac{V_R}{I} = \frac{100}{5} = 20\ \Omega$$

Both paths lead us to the same elegant truth. By placing a $20\ \Omega$ resistor in series, we perfectly balance the circuit and let the bulb shine safely!

Animated Solution for Physics - Current Electricity: An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance $R$ that must be put in series with the bulb, so that the bulb delivers 500 W is ...... $Ω$ .

Enter Numerical Value:

Visualized Solution

$B u l b S p ec i f i c a t i o n s$

$C a l c u l a t in g R_{b}$

$C i r c u i t A na l y s i s$

$V o l t a g e a cr oss R$

$F in d in g R$

$A l t er na t i v e M e t h o d$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Power of the Bulb

The Circuit Dilemma

The Elegant Conclusion

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bulb Specifications

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Rb​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Voltage across R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alternative Method