A heater is designed to operate with a power of `1000W` in a `100 V` line. It is connected in combination with a resistance of `10 Omega` and a resistance `R`, to a `100 V` mains as shown in figure. What will be the value of `R` so that the heater operates with a power of `62.5W`?

Options

A. `15Omega`
B. `10Omega`
C. `5Omega`
D. `25Omega`

Answer

C

Explanation

Potential difference across the `400Omega` resistance=30V. Therefore potential difference across the `300Omega` resitance=60-30V=30V. Let R be the resistance of the voltmeter. As the voltmeter is in the parallel with the `400Omega` their combined resistance is `R'=(400R)/((400+R))`
As the potential difference of 60V is equally shared between the `300Omega and 400Omega` resistance, R' should be equal to `300Omega`.Thus `300=(400R)/((400+R))`
Which gives,`R=1200Omega` which is the resistance of the voltmeter, is connected acrosst he `300Omega` resistance, their combined resistance is
`R''=(300R)/((300+R))=(300xx1200)/((300+1200))=240Omega`
Total resistance in the circuit=`400+240=640Omega`
Current in the circuit is `I=(60V)/(640Omega)=(3)/(32)A`
Voltmeter reading `="Potential difference across" 240Omega "resistance"`
`=(3)/(2)xx240=22.5V`