Test: Heating Effect of Electric Current - Class 10 MCQ

If the current passing through an electric heater has been halved then the heat produced will also be reduced to one fourth because H=I² RT.

Let n be the number of days

Here, work done is 36 x 10⁵ J = (18 J/s) x n x 1 x 60 x 60

Thus, n = 55.5 days

We know that, R = V² /P

Hence resistance of 1st bulb = (250)² / 200 = 625/2

resistance of 2nd bulb = (250)² / 100 = 625

Total resistance of the circuit when the bulbs are connceted in series = 625/2 + 625 = 1875/2 ohm

Therefore, total power consumed in the circuit , P = (250)² / (1875/2) = 67 watt

Electric power in terms of current = IR2 and in terms of voltage = V2/R and in terms of voltage and current both = VI

From ohm’s law, I = V/R = 220/440 = 0.5 A

H = I² × R × t = (0.5)² × 440 × 10 = 1100 J

E = P × t = 0.5 × 10 × 30 = 150 kWh
Cost of energy = 150 × 4 = Rs. 600

The Fuse wire connection are in between the appliance and main line. It is are very thin wire and when very high current passes through it suddenly, the fuse wire gets heated and melts due to the produced heat. This saves the household appliances

H = Pt

Voltage (V) = 270 Volts

Current (I) = 0.5 A

Power (P) = V x I
= 270 x 0.5 = 135 watt

P = I² R = (10)² × 5 = 500 watt = 0.5 kW

t = 30 minute = 0.5 hr

E = P × t = 0.5 kW × 0.5 hr = 0.25 kWh

Power = voltage * current 
=  Voltage * ( voltage / resistance) 
Hence, if P is power, V is Voltage and R is resistance, then R = V ^2/ R 
Hence for first bulb, R1 = V^2/75 = R
For second bulb, R2 = V^2 / 150
 = > ½ (V^2/75) = R/2 
=> R/2

Resistance of a wire is inversely proportional to the cross−sectional area:

The commercial unit of electrical energy is kilowatt hour.