Two heater `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, then total heat produced in 5 minutes will be

Options

A. `200 kJ`
B. `100 kJ`
C. `50 kJ`
D. `10 kJ`

Answer

B

Explanation

(b) For heater `A: 500 xx 10^(3) = (V^(2))/(R_(1)) (20 xx 60)`
Where `R_(1) = 100 Omega`
For heater `B`:
`100 xx 10^(3) = (V^(2))/(R_(2)) (10 xx 60)`
From (i) and (ii) `R_(2) = 25 Omega`
When heaters are connected in series :
`R_(eq) = R_(1) + R_(2)`
Hat produced :
`H = (V^(2))/(R_(1) + R_(2)) (5 xx 60)`
Form (i) and (iii) solve to get `H = 100 kJ`.