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  • Two heaters A and B are in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is <img alt="100\; \Omega" src="/late

Two heaters A and B are in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is 100\; \Omega If the same heaters are connected in series across the same voltage, the heat produced in 5 minutes will be

Option: 1

200 kJ 


Option: 2

100 kJ


Option: 3

50 kJ


Option: 4

10 kJ


Answers (1)

best_answer

\begin{aligned} & \frac{V^2}{100} \times 20 \times 60=500 \times 10^3 \quad \Rightarrow \quad V^2=\frac{5}{12} \times 10^5 \\ & \frac{V^2}{R_B} \times 10 \times 60=1000 \times 10^3 \quad \Rightarrow \quad R_B=25 \Omega \\ & \therefore \text { Heat in } 5 \mathrm{~min}=\left(\frac{\mathrm{V}^2}{\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{B}}}\right) \times 5 \times 60=100 \mathrm{~kJ} \\ & \end{aligned}

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Deependra Verma

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