A heater is designed to operate 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 mains as shown in figure. For the heater to operate at 62.5 W, the value of R should be ______ $\Omega$.

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Current Electricity

A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of

10Ω

and a resistance R, to a 100 V mains as shown in figure. For the heater to operate at 62.5 W, the value of R should be ______

Ω

Answer: 5

Heater's own resistance:

R_{h} = \frac{V ^{2}}{P} = \frac{10 0 ^{2}}{1000} = 10 Ω

.
For the heater to dissipate 62.5 W at

R_{h} = 10 Ω

V_{h} = P R_{h} = 625 = 25

V, so

i_{H} = V_{h} / R_{h} = 2.5

A.
The fixed

10 Ω

resistor (in series with the parallel heater–

R

combination) carries the total current

i_{1}

and drops the remaining

100 - 25 = 75

i_{1} = 75/10 = 7.5

A.
Current through

R

i_{R} = i_{1} - i_{H} = 7.5 - 2.5 = 5

R = \frac{V _{h}}{i _{R}} = \frac{25}{5} = 5 Ω

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer