Advertisements
Advertisements
Question
Three 250 W heaters are connected in parallel to a 100 V supply, Calculate the resistance of each heater.
Advertisements
Solution

The resistance of each heater
Since P = `"V"^2/"R"`
or I = `"V"^2/"R" = (100 xx 100)/250 = 40 Omega`
Therefore, 40 Ω is the resistance of each heater.
RELATED QUESTIONS
The unit for expressing electric power is:
(a) volt
(b) joule
(c) coulomb
(d) watt
When a current I flows through a resistance R for time t, the electrical energy spent is given by:
(a) IRt (b) I²Rt (c) `IR^2t` (d) `I^2R ⁄t`
An electric bulb is rated at 220 V, 100 W. (a) what is its resistance? (b) what safe current can be passed through it?
Two resistors A and B of resistance 4 Ω and 6 Ω respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, (ii) the power dissipated in each resistor.
Two resistors A and B of 4 Ω and 6 Ω respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate the power supplied by the battery.
I ampere of current is passed through a resistor of resistance R ohm for time t second. Obtain expression for
(i) the electrical energy, and
(ii) the electrical power spent.
Find the heat produced in joules when a current of 10 amperes flows through a connecting wire of resistance 2 × 10-2 ohms and heating element of resistance 250 ohms for 2 seconds.
The resistance of filament of an electric heater is 500 Ω It is operated at 200V for 1 hour daily. Calculate the current drawn by the heater and the energy consumed in kWh, by the heater in a month of 30 days.
An electrical appliance is rated 1500 W, 250 V. This appliance is connected to 250 V mains. Calculate:
(i) the current drawn,
(ii) the electrical energy consumed in 60 hours,
(iii) the cost of electrical energy consumed at Rs. 2.50 per kWh.
