Q11

1 markMCQSection A

Two heaters rated as $(P_1, V)$ and $(P_2, V)$ are connected in series across a dc source of $V/2$ volt. The power consumed by the combination will be −

Q: Two heaters rated as $(P_1, V)$ and $(P_2, V)$ are connected in series across a dc source of $V/2$ volt. The power consumed by the combination will be −

**Correct option: (D) $P_1P_2 / [4(P_1 + P_2)]$** Step-wise: - Resistances from ratings: $R_1 = V^2/P_1$, $R_2 = V^2/P_2$. - Series total: $R = R_1 + R_2 = V^2(P_1 + P_2)/(P_1P_2)$. - Power at supply V/2: $P = (V/2)^2/R = (V^2/4) \cdot P_1P_2 / [V^2(P_1 + P_2)]$ = **$P_1P_2 / [4(P_1 + P_2)]$**.

Current Electricity

Electric Power — Heaters in Series

Options

(A)

(P_1 + P_2)

(B)

\frac{P_1 + P_2}{2}

(C)

\frac{P_1 P_2}{2(P_1 + P_2)}

(D)

\frac{P_1 P_2}{4(P_1 + P_2)}

Official Answer

Correct option: (D) $P_1P_2 / [4(P_1 + P_2)]$

Step-wise:

Resistances from ratings: $R_1 = V^2/P_1$ , $R_2 = V^2/P_2$ .
Series total: $R = R_1 + R_2 = V^2(P_1 + P_2)/(P_1P_2)$ .
Power at supply V/2: $P = (V/2)^2/R = (V^2/4) \cdot P_1P_2 / [V^2(P_1 + P_2)]$ = $P_1P_2 / [4(P_1 + P_2)]$ .

electric powerheaters in seriesR equals V squared by Pseries combinationpower ratingP1P2 by 4(P1+P2)current electricityhalf supply voltage

Marking Scheme

11 mark: correct option (D) $P_1P_2/[4(P_1+P_2)]$ .
2Key reasoning: $R = V^2/P$ for each heater, series resistances add, and applied voltage $V/2$ introduces a factor of $\frac{1}{4}$ .

Hint

Convert ratings to resistances ( $R = V^2/P$ ), add them in series, then use $P = (V/2)^2/R$ .

Quick Oral Answer

Each heater's rating fixes its resistance as $V^2/P$ ; adding them in series and applying half the rated voltage gives a total power of $P_1P_2$ divided by four times the sum of the powers.

Analysis & Explanation

This combines the power rating of an appliance with the series-resistance rule, then re-computes power at a different supply voltage.

Concept: A rating $(P, V)$ fixes the resistance, not the power: $R = V^2/P$ . Power actually dissipated depends on the applied voltage through $P = V^2_{\text{applied}}/R$ .

Working:

$R_1 = V^2/P_1$ and $R_2 = V^2/P_2$ .
In series $R = R_1 + R_2 = V^2/P_1 + V^2/P_2 = V^2(P_1 + P_2)/(P_1P_2)$ .
Supply voltage is V/2, so total power $P = (V/2)^2/R$ .
$P = (V^2/4) \div [V^2(P_1 + P_2)/(P_1P_2)] = P_1P_2/[4(P_1 + P_2)]$ .

Why (D) is correct: it accounts for both the series combination (product-over-sum resistance behaviour) and the halved supply voltage (factor $\frac{1}{4}$ in power).

Why the others are wrong:

(A) $P_1+P_2$ and (B) $\frac{P_1+P_2}{2}$ are what you would (incorrectly) get by adding powers, valid only for a parallel connection at the rated voltage V.
(C) $\frac{P_1P_2}{2(P_1+P_2)}$ is the power for the series combination at the full rated voltage V; it forgets the $\frac{1}{4}$ factor from using V/2.

Exam trap: Powers add only in parallel; in series it is the resistances that add, and the voltage here is halved.

Common Mistakes

1Adding the powers $(P_1 + P_2)$ as if they combine directly — powers add only in parallel at the rated voltage, not in series.
2Forgetting the supply is $V/2$ , so omitting the factor $\frac{1}{4}$ and getting option (C) instead of (D).
3Treating the rating power as the power actually consumed, ignoring that resistance is the fixed quantity: $R = V^2/P$ .

Interesting Facts

This is why two identical heaters in series glow far dimmer than a single one: doubling resistance and (here) halving voltage cuts the total power drastically.

For appliances, only the resistance is physically fixed by the element; the labelled wattage is merely the power at the labelled voltage.

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Frequently Asked Questions

Why do the resistances add but the powers do not, in a series connection?

In series the same current flows and voltages add, so resistances add directly. Powers do not simply add because the power in each element depends on how the supply voltage divides across it, which changes once they are combined.

Where does the factor of $\frac{1}{4}$ come from?

Power is proportional to the square of the applied voltage ( $P = V^2/R$ ). Since the supply is $V/2$ instead of V, the power is scaled by $(1/2)^2 = \frac{1}{4}$ .