Question

n identical light bulbs, each designed to draw $P$ power 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. $\dfrac{P}{n}$
D. $\dfrac{P}{n^2}$

Answer 1

Hint: Power of each bulb is given by $P = \dfrac{{{V^2}}}{R}$ where $V$ is the voltage rating of the bulb and $R$ is its resistance.
The equivalent resistance in a series combination of n resistors is given by ${R_{eq}} = {R_1} + {R_2} + {R_3} + ..... + {R_n}$
Current through each bulb in series connection is same and given by $I = \dfrac{V}{{{R_{eq}}}}$ and the total power drawn can be calculated as ${P_t} = {I^2}{R_{eq.}}$

Complete step by step solution:
Electric power can be defined as the rate at which energy is transferred to an electric circuit. The source of energy can be a battery or a circuit element like a resistor that can release energy in the form of heat. For any circuit element, the power is equal to the potential difference across the element multiplied by the current. By Ohm's Law, V = IR, and so there can be additional forms of the electric power formula for resistors.
We first calculate the resistance of each bulb. For this, we have
The power of each bulb is given by $P = \dfrac{{{V^2}}}{R}$ where $V$ is the voltage rating of the bulb and $R$ is its resistance.
So, the resistance of each bulb is given by $R = \dfrac{{{V^2}}}{P}$
Now as given in the question that there are \[n\] identical bulbs which are joined in series.
So, as we know that the equivalent resistance in a series combination of n resistors is given by ${R_{eq}} = {R_1} + {R_2} + {R_3} + ..... + {R_n}$
So, equivalent resistance of the bulbs will be ${R_{eq}} = nR$
Now, current through each bulb in series connection is same and given by $I = \dfrac{V}{{{R_{eq}}}}$ and the total power drawn can be calculated as ${P_t} = {I^2}{R_{eq}}$.
So, $I = \dfrac{V}{{nR}}$ and the total power is drawn is given by
${P_t} = {\left( {\dfrac{V}{{nR}}} \right)^2}\left( {nR} \right) = \dfrac{{{V^2}}}{{nR}}$
On simplification we have ${P_t} = \dfrac{P}{n}$

The total power drawn by the given circuit is $\dfrac{P}{n}$. Hence, option C is correct.

Note:
When the resistors are connected in series, the current must flow through circuit components in sequence. Since the current has to pass through each resistor sequentially through the circuit, the total resistance in the circuit is equal to the sum of each resistances. That’s why current through each bulb in series connection is the same.