A 100 W bulb B1, and two 60 W bulbs B2 and B3, are connected to a 250 V source, as shown in the figu

Q: Why do we assume the voltage rating is the same for all bulbs?

In standard textbook problems, unless specified otherwise, bulbs are assumed to be rated for the standard household voltage. Their relative resistances depend only on their power ratings.

Q: Why does the 100W bulb consume less power than the 60W bulb in series?

A 100W bulb is designed to draw more current when connected to full voltage, which means it has a lower resistance. When forced into a series circuit, the current is the same for both, so the lower resistance bulb dissipates less power.

Q: Can we calculate the exact power without knowing the rated voltage?

We cannot calculate the exact numerical value of power without the rated voltage, but we can perfectly determine the ratio and order of power dissipated because the rated voltage term cancels out when comparing.

Q: Why is the voltage across $B_2$ less than 250V?

$B_1$ and $B_2$ are in series across the 250V source. According to Kirchhoff's Voltage Law, the sum of the voltage drops across $B_1$ and $B_2$ must equal 250V. Thus, $B_2$ only gets a fraction of the total voltage.

Q: What if $B_1$ and $B_2$ were in parallel?

If all three were in parallel, they would all receive the full 250V. In that case, their output power would exactly match their rated power, meaning $W_1 = 100\text{ W}$, and $W_2 = W_3 = 60\text{ W}$.

Question

A 100 W bulb $B_1$, and two 60 W bulbs $B_2$ and $B_3$, are connected to a 250 V source, as shown in the figure. Now $W_1, W_2$ and $W_3$ are the output powers of the bulbs $B_1, B_2$ and $B_3$ respectively. Then, Options: (A) $W_1 > W_2 = W_3$ (B) $W_1 > W_2 > W_3$ (C) $W_1 < W_2 = W_3$ (D) $W_1 < W_2 < W_3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a household circuit. We have three bulbs: $B_1$ rated at $100	ext{ W}$, and $B_2$ and $B_3$ both rated at $60	ext{ W}$. They are connected to a $250	ext{ V}$ source.

Looking at the circuit diagram, we can see that $B_1$ and $B_2$ are connected in series on the top branch. Meanwhile, $B_3$ is sitting on its own parallel branch, connected directly across the voltage source.

Our goal is to find the actual power dissipated by each bulb, denoted as $W_1$, $W_2$, and $W_3$, and arrange them in increasing order.

### The Master Equation

To understand how much power a bulb will actually consume in a circuit, we first need to know its intrinsic property: **resistance**.

The resistance of a bulb is determined by its manufacturer's power rating and voltage rating. The master equation connecting these is:
$$P = \frac{V^2}{R} \implies R = \frac{V^2}{P}$$

Assuming all bulbs are designed for the same standard voltage $V$, we can write their resistances as:
$$R_1 = \frac{V^2}{100}$$
$$R_2 = \frac{V^2}{60}$$
$$R_3 = \frac{V^2}{60}$$

Notice a beautiful counter-intuitive fact here: **a higher power rating means a lower resistance**. Since $100 > 60$, it mathematically follows that $R_1 < R_2$. Also, since $B_2$ and $B_3$ have the same rating, $R_2 = R_3$.

### Power in Series

Now, let's focus on the top branch where $B_1$ and $B_2$ are in series.

In a series circuit, the current $I$ is the exact same for all components. The power consumed by a resistor in series is best analyzed using the formula:
$$W = I^2 R$$

Since the same current flows through both $B_1$ and $B_2$, the power they dissipate is directly proportional to their resistance. We already established that $R_1 < R_2$. Therefore, the $100	ext{ W}$ bulb actually consumes less power than the $60	ext{ W}$ bulb in this setup!
$$W_1 < W_2$$

### Power in Parallel

What about $B_3$? It is connected in parallel with the source, meaning it gets the full $250	ext{ V}$ across its terminals.

Its power output is simply:
$$W_3 = \frac{250^2}{R_3}$$

Now, let's compare this to $B_2$. Bulb $B_2$ is in series with $B_1$, which means they have to share the $250	ext{ V}$ source. The voltage across $B_2$, let's call it $V_2$, is strictly less than $250	ext{ V}$.

The power consumed by $B_2$ can be written as:
$$W_2 = \frac{V_2^2}{R_2}$$

Since $V_2 < 250	ext{ V}$, it is obvious that $\frac{V_2^2}{R_2} < \frac{250^2}{R_2}$. And since $R_2 = R_3$, we can confidently say:
$$W_2 < W_3$$

### Final Calculation

We have successfully decoded the physics of the circuit.

From the series branch, we found $W_1 < W_2$. From comparing the parallel branches, we found $W_2 < W_3$.

Combining these two inequalities gives us our final, elegant result:
$$W_1 < W_2 < W_3$$

This perfectly matches option (d).

Animated Solution for Physics - Current Electricity: A 100 W bulb $B_{1}$ , and two 60 W bulbs $B_{2}$ and $B_{3}$ , are connected to a 250 V source, as shown in the figure. Now $W_{1}, W_{2}$ and $W_{3}$ are the output powers of the bulbs $B_{1}, B_{2}$ and $B_{3}$ respectively. Then,

Select Answer:

Visualized Solution

Circuit Analysis

Resistance Formula

Calculating Resistances

Comparing $R_{1}$ and $R_{2}$

Power in Series Branch

Power of $B_{3}$

Comparing $W_{2}$ and $W_{3}$

Final Conclusion

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

Analyzing the Setup

The Master Equation

Power in Series

Power in Parallel

Final Calculation

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