Question

n identical light bulbs, each designed to draw power \( P \) from a certain voltage supply are joined in series across that supply. The total power which they will draw is:

• A. \( nP \)
• B. \( P \)
• C. \( \frac{P}{n} \)
• D. \( Pn^{-2} \)

Hint:

In a series circuit, the total power is the sum of the powers dissipated across all resistors (light bulbs in this case).

Answer 1

The Correct Option is C

To determine the total power drawn by \( n \) identical light bulbs connected in series, let's consider the following:

• Each bulb has a power rating of \( P \), which means when it is connected individually to the supply voltage, it would draw power \( P \).
• When bulbs are connected in series, the total voltage is divided among them.
• If \( V \) is the total voltage of the source, each bulb will receive a voltage of \( \frac{V}{n} \) because the total voltage is shared equally among the bulbs in series. 
• The power drawn by each bulb when connected in series is given by the formula for power: \( P = \frac{V^2}{R} \), where \( R \) is the resistance of one bulb.
• In series, the voltage across each bulb is \( \frac{V}{n} \), so the power across each bulb is:

\[\text{Power per bulb} = \left(\frac{V}{n}\right)^2 \times R\]

• The total power drawn by all \( n \) bulbs is:

\[\text{Total power} = n \times \left(\frac{V}{n}\right)^2 \times R = \frac{nV^2}{n^2R} = \frac{V^2}{nR}\]

• Since each bulb alone draws power \( P = \frac{V^2}{R} \), the expression for total power becomes:

\[\text{Total power} = \frac{P}{n}\]

Therefore, the total power drawn by the bulbs when connected in series is \( \frac{P}{n} \).

Answer 2

In a series circuit, the current through each light bulb is the same. The total resistance \( R_{\text{total}} \) is the sum of the individual resistances \( R_{\text{individual}} \) of each bulb. The power drawn by each bulb is \( P = \frac{V^2}{R_{\text{individual}}} \), and the total power is: \[ P_{\text{total}} = \frac{V^2}{R_{\text{total}}} \] Since the total resistance in series is \( R_{\text{total}} = nR_{\text{individual}} \), the total power becomes: \[ P_{\text{total}} = \frac{P}{n} \]

Answer 3

Each bulb is rated to draw power \(P\) when it gets its full rated voltage \(V\) all to itself, which means each bulb draws rated current \(I_0 = \dfrac{P}{V}\) and has resistance \(R = \dfrac{V}{I_0} = \dfrac{V^2}{P}\).

When \(n\) identical bulbs are wired in series across the same supply voltage \(V\), the same single current \(I\) flows through every bulb, and the total resistance the supply sees is \(n\) times one bulb's resistance, \(R_{\text{total}} = nR\). By Ohm's law, the current that actually flows is

\[ I = \frac{V}{nR} = \frac{V}{n\left(\dfrac{V^2}{P}\right)} = \frac{P}{nV} \]

The total power delivered by the supply is \(V\) times this current:

\[ P_{\text{total}} = VI = V\cdot\frac{P}{nV} = \frac{P}{n} \]

1. Option 1 (\(nP\)): this would be the power drawn if all \(n\) bulbs were each independently connected across the full voltage \(V\), as in a parallel arrangement, not squeezed into one series loop sharing a reduced current.
2. Option 2 (\(P\)): would only hold if adding more bulbs in series left the current, and hence the power, unchanged, but extra series resistance always reduces the current below its single-bulb value.
3. Option 3 (\(\frac{P}{n}\)): matches the value obtained from the Ohm's-law calculation above.
4. Option 4 (\(Pn^{-2}\)): falls off faster with \(n\) than the \(1/n\) relationship the series circuit actually produces.

So the correct answer is \(\dfrac{P}{n}\).