Cell in electric circuit is a source of energy. Cell has two terminals one is of positive voltage and another is of negative voltage. Hence it provides potential difference. potential difference means voltage in a battery is due to chemical energy.
Consider a electric circuit as shown in figure.
The resistor of resistance R is connected in series to a battery with the help of a key. The ammeter is connected in series to calculate current in the circuit. Also we connected voltmeter across the resistor R to measure the voltage drop by resistor. Battery is continuously providing voltage or energy. where does this energy go? The applied energy will divide into
1) Resistor.
2) wire.
3) Battery.
1) Resistor
When current is passing through the circuit, some equipment get heat up. Did you touch the fan when it off. What caused the fan heat up? When electron travel in a circuit, it collides with other ions and due to continuously collision the heat is created in the circuit.
Suppose the current I will flow in the resistor R due to applied voltage V.
Now, We calculate the heat created in the circuit.
As we know that,
Power = Energy / time.
So electric power in a circuit is
Power = Electrical energy/ flow of charge. ——-1
As, voltage = energy / charge
⇒ energy = voltage x charge.
Put this value in equation 1
Power = ( Voltage x charge)/ time
But charge / time = current
There for,
Power = current × voltage
P = I × V——————-2
This is the formula for power.
Due to that power heat energy will produce in the circuit,
Power = Energy / time
Put this in equation 2,
E/t. = I ×V
E = I × V t
Energy will be in the form of heat.
Heat = I × V × t——-3
Also we put value of voltage in the form of resistor, by Ohm’s law
V = I × R , put this in equation 3
Heat = I² × R × t
This equation is called as joule’s law or joule’s heating law.
Statement:
Boyle’s law state that the heat produced in the resistor connected in circuit is directly proportional to
- Square of current flowing through resistor
- Resistance of resistor.
- Time of flowing current in the circuit.
Q) Calculate heat produced in the bulb if a 2A current is flowing in it for one hour and resistor of the bulb is 60 ohm.
Solution:
Given, I = 2 A,. R = 60 ohm, t = 1 hour = 3600 sec
As, H = I² × R × t
H = 2² × 60 × 3600
= 4 × 60 ×3600
H = 864000 J
The heat produced in the circuit is 8.64 x 10⁵ J
Q) When we apply 220 volt to a bulb of ratting 60 watt then Resistance and current flowing in that bulb.
Q) It the heater produced 200 J heat each second in 50 ohm resistance then calculate applied potential difference.
Practical applications of heating effect:
We know that, when current is passing through the circuit then the energy convert into heat. The heat generated in circuit has so many practical applications.
1) Electric Bulb:
Electric bulb convert electrical energy into light energy. When we pass current into the bulb, the bulb heat up and due to heating it emit radiations.
2) Electric Fuse:
Fuse protect electrical appliances. The fuse connected in series with electrical appliances. At desired value of current, fuse flow current in it. But when current exceed the value of limit in circuit then due to heating fuse burn out and circuit break. Hence our electric appliance protect from damage.
3) Electric iron
4) Electric heater
5) Electric oven
6) Electric toaster
7) Thermistor:
It is. Type of resistor whose resistance depend on temperature.
Q) Write down the applications of joule’s heating effect.