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

Options

A. `10 Omega`
B. `62.5 Omega`
C. `(1)/(5) Omega`
D. `5 Omega`

Answer

D

Explanation

Resistance of heater `R_(h)= (V^(2))/( P)= (100xx 100)/( 1000)= 10 Omega`
`therefore` total resistance of circuit `=10 + (10 xx R)/( 10+ R)= (100+ 20 R)/(10 + R)`
Current in heater `= I. (R )/( 10+ R)`
`= (100)/( (100+20 R)/( 10 + R))xx (R )/(10+R)= (5 xx R)/( 5 + R)`
`therefore` Power `P= I^(2) R_(h) rArr 62.5 = ((5R)/( 5+ R))^(2) xx 10 therefore R= 5 Omega`