If the current passing through a heater is halved, then the heat produced by it becomes
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.
An electric bulb is marked 18watt-240 volt. If it is used across 240 V power line for one hour daily, calculate the number of days to consume 1 unit of electric energy.
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
Two bulbs marked 200 watt-250 volts and 100 watt-250 volts are joined in series to 250 volts supply. Power consumed in circuit is
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
Which of the following terms does not represent electrical power in a circuit ?
Electric power in terms of current = IR2 and in terms of voltage = V2/R and in terms of voltage and current both = VI
The resistance of a conductor increases with-
I: Increase in length
II: Increase in volume
III: Decrease in area
How much heat energy is produced in 10 second in an electric iron having resistance 440 Ω, when a potential difference of 220 volt is applied across it?
From ohm’s law, I = V/R = 220/440 = 0.5 A
H = I2 × R × t = (0.5)2 × 440 × 10 = 1100 J
An electric refrigerator rated 500 W operates 10 hr per day. What is the cost of the energy to operate it for 30 days at Rs. 4.00 per kWh ?
E = P × t = 0.5 × 10 × 30 = 150 kWh
Cost of energy = 150 × 4 = Rs. 600
Which property of the electricity is responsible for use of Fuse wire in household wiring?
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
Two light bulbs P and Q are identical in all respects, except that P’s filament is thicker than Q’s. If the same potential difference is applied to each, then
Find the potential difference across the resistor, when 125 J of heat is produced each second in a 5 ohm resistor.
H = Pt
A bulb is rated as 270V, 0.5 A. Its power is
Voltage (V) = 270 Volts
Current (I) = 0.5 A
Power (P) = V x I
= 270 x 0.5 = 135 watt
How much energy is transferred when 10 A current flows through a resistor of 5 ohm for 30 minutes ?
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
Two light bulbs are marked 230 V; 75 W and 230 V; 150 W. If the first bulb has a resistance R, then the resistance of the second is;
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
A certain wire has a resistance R. The resistance of another wire identical with the first and having twice its diameter is;
Resistance of a wire is inversely proportional to the cross−sectional area:
Commercial unit of electric energy is
The commercial unit of electrical energy is kilowatt hour.