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A 25 watt, 220 volt bulb and a 100 watt, 220 volt bulb are connected in series across 440 volt lines.(A) only 100 watt bulbs will fuse(B) only 25 watt bulbs will fuse(C) none of these bulbs will fuse(D) both bulbs will fuse.

Last updated date: 11th Aug 2024
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Hint: When bulbs are connected in such a manner that one terminal of one bulb is kept open and another terminal is connected to one terminal of another bulb then they are considered to be connected as in series combination. Formula used to solve the problem: $P = \dfrac{{{V^2}}}{R}$ and $P = {I^2}R$ where P is the power, R is resistance and V is the Voltage or potential applied across the bulb.

Complete step by step solution:
Given
Calculate the resistance of each bulb
Since, Power $P = \dfrac{{{V^2}}}{R}$
$\Rightarrow R = \dfrac{{{V^2}}}{P}$
So, ${R_1} = \dfrac{{{V^2}}}{{{P_1}}} = \dfrac{{{V^2}}}{{25}}$
and ${R_2} = \dfrac{{{V^2}}}{{{P_2}}} = \dfrac{{{V^2}}}{{100}}$
Calculate the equivalent resistance
$R{}_{eq} = {R_1} + {R_2}$
$\Rightarrow R{}_{eq} = {V^2}\left( {\dfrac{1}{2} + \dfrac{1}{{100}}} \right)$
$\therefore R{}_{eq} = \dfrac{{{V^2}}}{{20}}$
Since, In series current is constant
So, when connected to supply of voltage 440 v the current in the circuit would be
$\Rightarrow I = \dfrac{{V'}}{{R{}_{eq}}}$
Since, V’ = 2V
$\Rightarrow I = \dfrac{2V}{\dfrac{V^2}{20}}$
$\therefore I = \dfrac{{40}}{{{V_0}}}$
Now
Power generated in the bulb 1 will be
$\Rightarrow P{}_1 = {I^2}{R_1} = {\left( {\dfrac{{40}}{V}} \right)^2} \times \left( {\dfrac{{{V^2}}}{{25}}} \right)$
$\therefore P{}_1 = 64W$
Power generated in the bulb 2 will be
$\Rightarrow P{}_2 = {I^2}{R_2} = {\left( {\dfrac{{40}}{V}} \right)^2} \times \left( {\dfrac{{{V^2}}}{{100}}} \right)$
$\therefore P{}_2 = 16W$
Here it is clearly seen that
$64W>25W$
Therefore, Only bulb 1 or of 25 watt will get fused.

Hence option (B) is the correct answer.

Note: Any resistive electronic device consumes electrical power and it does not depend upon the direction of current.
In both the above cases, power is only consumed and this power consumed is given by

$P = \dfrac{{{V^2}}}{R} = {I^2}R = VI$
In any electrical circuit, law of conservation of energy is followed. i.e.
Net power supplied by all the batteries of the circuit =net power consumed by all the resistors in the circuit.