Question

Two equal resistances when connected in series to a battery consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be
1. 60 W
2. 30 W
3. 240 W
4. 120 W

Answer 1

Correct Answer - Option 3 : 240 W

Concept:

Given,

Power = 60 W

Let P₁ and P₂ be the individual electric powers of the two resistances, respectively.

In series combination, power is

\({P_0} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = 60\;{\rm{W}}\) 

Since, the resistances are equal and the current through each resistor in series combination is also same.

Then,

P₁ = P₂

Calculation:

Given

\(\frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = 60\;{\rm{W}}\) 

\(\frac{{{P_1}{P_1}}}{{{P_1} + {P_1}}} = 60\;W\) 

\(\frac{{P_1^2}}{{2{P_1}}} = 60\;W\) 

\(\frac{{{P_1}}}{2} = 60\;W\) 

P₁ = 120 W

Thus,

P₁ = P₂ = 120 W

In parallel combination, power is

P = P₁ + P₂

P = 120 + 120 = 240 W

Thus, if these resistances are connected in parallel combination to the battery, then the electric power consumed will be 240 W

Alternate method

Let R be the resistance

∴ Net resistance in series = R + R = 2R

\(P = \frac{{{V^2}}}{R}\left[\because {R = 2R} \right]\) 

\(P = \frac{{{V^2}}}{{2R}} = 60\;W\) 

\(\frac{{{V^2}}}{R} = 120\;W\) 

New resistance in parallel

\( = \frac{{R \times R}}{{R + R}} = \frac{{{R^2}}}{{2R}} = \frac{R}{2}\) 

\(P' = \frac{{{V^2}}}{R}\) 

\(P' = \frac{{{V^2}}}{{\frac{R}{2}}}\) 

\(P' = 2\left( {\frac{{{V^2}}}{R}} \right)\) 

P' = 2(120) = 240 W

Thus, if these resistances are connected in parallel combination to the battery, then the electric power consumed will be 240 W.