Two heating wire of equal length are first connected in series and then in parallel to a constant voltage source. The rate of heat produced in two cases is (parallel to series)

1 : 4

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4 : 1

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1 : 2

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2 : 1

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Solution

The correct option is B $4 : 1$
Let $R$ be the resistance of each wire.
In series: Equivalent resistance , $R_{S} = R + R = 2 R$

Heat per second $=$ Power $= P_{S} = \frac{V^{2}}{R_{S}} = \frac{V^{2}}{2 R}$

In parallel: Equivalent resistance , $R_{P} = \frac{R R}{R + R} = R / 2$

Heat per second $=$ Power $= P_{P} = \frac{V^{2}}{R_{P}} = \frac{V^{2}}{R / 2} = \frac{2 V^{2}}{R}$
Thus, $\frac{P_{P}}{P_{S}} = \frac{2 V^{2}}{R} \times \frac{2 R}{V^{2}} = 4$

$P_{P} : P_{S} = 4 : 1$

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Similar questions

Q. Two identical heater wires are first connected in series and then in parallel with a source of electricity. The ratio of heat produced in the two cases ?

Two conducting wires of the same material and of equal length and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-

(a) 1:2

(b) 2:1

(d) 4:1

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Two heating wire of equal length are first connected in series and then in parallel to a constant voltage source. The rate of heat produced in two cases is (parallel to series)

The correct option is B 4:1Let R be the resistance of each wire.In series: Equivalent resistance ,RS=R+R=2RHeat per second =Power =PS=V2RS=V22RIn parallel: Equivalent resistance ,RP=RRR+R=R/2Heat per second = Power=PP=V2RP=V2R/2=2V2RThus, PPPS=2V2R×2RV2=4PP:PS=4:1