In the previous section we completed a discussion on the 'relation between electricity and magnetism'. In this chapter, we will see the 'effects of electric current'.
Let us first see how an electric current can produce a 'heating effect'. We will do an activity:
1. In the fig.10.1 below, A and B are two beakers of 200 mL capacity.
• Each beaker contains 100 mL of water.
• The conductor PQ in the beaker A is a nichrome wire
• The conductor RS in the beaker B is a copper wire
• Both the conductors have the same length and diameter
2. Trial 1:
Measure the temperature of the water in both the beakers and note it down. Now we can start the trials:
• Turn the switch on
• Note down the current shown by the ammeter reading
• Allow the current to flow through the circuit for three or four minutes
• Turn off the switch and immediately measure the temperature of water in each beaker.
♦ Let them be tA and tB
• We will see that tA is larger than tB
3. So the observations are over. Let us analyse them:
• Beakers A and B are connected in series. So the same current (indicated by the ammeter) will be flowing through both nichrome and copper
• Both nichrome and copper pieces are of the same size. Because they have the same length and diameter
• Even when current, lengths and diameters are the same, nichrome produced greater heat. Why is that so?
Ans: Nichrome offers greater resistance to the flow of electricity. That is., electricity cannot flow easily through nichrome. So most of the electrical energy will be converted into heat energy.
4. But the opposite happens in the case of copper
• Copper offers lesser resistance to the flow of electricity. That is., electricity can flow easily through copper. So only a lesser electrical energy will be converted into heat energy.
• Since in this trial, current, lengths and diameters of both nichrome and copper are the same, we can say that, resistance to flow of electricity depends on the material with which the conductor is made. We discussed it in a previous chapter here.
■ So we can write: Greater the resistance (R), greater is the heat energy produced
5. Trial 2:
• Allow the water in the beakers to cool down to room temperature.
• Turn on the switch and pass current for five minutes
• Note down the current. Let it be I1
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t1
6. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for seven minutes
• The current must be the same I1 as before. Ensure this by the ammeter reading
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t2
• We will see that, t2 is greater than t1
■ So we can write: Greater the 'time duration (t) for which current is passed through a conductor', greater is the heat energy produced
7. Trial 3:
• Allow the water to cool down to room temperature. Measure this temperature and note it down
• Turn the switch on and allow the current to pass for three minutes
• Note down the current I1
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t1
8. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for the same three minutes
• The current I1 this time must be greater than I1.
♦ Ensure this by the ammeter reading. The current can be increased by adjusting the rheostat
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t2
We will see that t2 is greater than t1
■ So we can write: Greater the 'current (I) which is passed through a conductor', greater is the heat energy produced
From the above activity, we can write:
The heat energy produced in a conductor due to the flow of current through it, depends on three items:
(i) The resistance (R) of the conductor
(ii) The time duration (t) for which the current is passed through the conductor
(iii) The intensity of the current (I)
• The English scientist James Prescott Joule discovered the relation connecting those three factors. The relation is known as Joule's Law:
■ The heat generated in a current carrying conductor is the product of the square of the current (I) in the conductor, the resistance (R) of the conductor and the time (t) of the flow of current
In simple terms, we can say:
The heat will be equal to the product of three items:
(i) Square of the current (I2)
(ii) Resistance (R)
(iii) time (t)
• So if H is the heat generated, then:
Eq.10.1: H = I2Rt
• I is measured in ampere, R in ohm and t in seconds
Solution:
• We have: H = I2Rt
• Case 1: H1 = (I1)2Rt
• Case 2: Let H2 be the heat generated when a current I2 passes for the same t seconds
• Then we can write:
H2 = (I2)2Rt
• Taking ratios, we get: H1⁄H2= (I1)2Rt⁄(I2)2Rt = (I1)2⁄(I2)2 = [(I1)⁄(I2)]2
But given that I2 = 2I1
• So we get: H1⁄H2 = [(I1)⁄(2I1)]2 = [(1)⁄(2)]2 = [1⁄4]
⟹ H1⁄H2 = 1⁄4 ⟹ H2 = 4H1
• So we can write: The new heat will be four times the original heat
■ Heat of H1 joules is generated in a conductor when a current I1 passes through it for t seconds. How much heat will be generated if the current is halved and is passed for the same t seconds?
Solution:
• We have: H = I2Rt
• Case 1: H1 = (I1)2Rt
• Case 2: Let H2 be the heat generated when a current I2 passes for t seconds
• Then we can write:
H2 = (I2)2Rt
• Taking ratios, we get: H1⁄H2= (I1)2Rt⁄(I2)2Rt = (I1)2⁄(I2)2 = [(I1)⁄(I2)]2
But given that I2 = 0.5I1
• So we get: H1⁄H2 = [(I1)⁄(0.5I1)]2 = [(1)⁄(0.5)]2 = [(1)⁄(5/10)]2 = [(10)⁄(5)]2 = [2]2 = 4
⟹ H1⁄H2 = 4 ⟹ H2 = H1⁄4
• So we can write: The new heat will be one fourth of the original heat
• The equation H = I2Rt gives us the relation between heat and the quantities: [current, resistance, time]
• Can we relate heat to voltage? Let us try:
• From ohm's law, we have: V = IR (Details here)
• From this we get: I = V⁄R
• So we can use (V⁄R) instead of I
• Thus we get: H = (V⁄R)2Rt ⟹ H = (V2⁄R2)Rt = (V2⁄R)t = V2t⁄R
So we can write:
Eq.10.2: H = V2t⁄R
Another derivation:
• From ohm's law, we have: V = IR
• From this we get: R = V⁄I
• So we can use (V⁄I) instead of R
• Thus we get: H = I2(V⁄I)t ⟹ H = VIt
So we can write:
Eq.10.3: H = VIt
Now we will see a solved example:
Solved example 10.1:
A bulb of resistance 920 Ω works on 230 V supply. Calculate the quantity of heat generated in 3 minutes
Solution:
1. Given: R = 920 Ω, V = 230 volts, t = 3 minutes = (3 × 60) = 180 seconds
2. We want a relation that will connect the heat (H) to: [R, V and t]
• So we will use Eq.10.2: H = V2t⁄R
3. Substituting the values, we get: H = (230)2×180⁄920 = 10350 J
Another method:
1. We have the basic equation: H = I2Rt
2. But the current I is not given. We can calculate it using the Ohm's law: V = IR
• So we get: I = V⁄R = 230⁄920 = 1⁄4 = 0.25 A
3. Thus we get: H = I2Rt = H = (0.25)2 × 920 × 180 = 10350 J
One more method:
1. Here we will try to use Eq.10.3: H = VIt
• We have calculated I in the previous method. So we can use it
2. Substituting the values, we get: H = 230 × 0.25 × 180 = 10350 J
Solved example 10.2
An electric iron works on 230 V. A current of 3 A flows through it for half an hour. Calculate the amount of heat energy generated
Solution:
1. Given: V = 230 volts, I = 3 A, t = 30 minutes = (30 × 60) = 1800 seconds
2. We want a relation that will connect the heat (H) to: [V, I and t]
• So we will use Eq.10.3: H = VIt
3. Substituting the values, we get: H = 230 × 3 × 1800 = 1242000 J = 1242 kJ
Let us first see how an electric current can produce a 'heating effect'. We will do an activity:
1. In the fig.10.1 below, A and B are two beakers of 200 mL capacity.
 |
| Fig.10.1 |
• The conductor PQ in the beaker A is a nichrome wire
• The conductor RS in the beaker B is a copper wire
• Both the conductors have the same length and diameter
2. Trial 1:
Measure the temperature of the water in both the beakers and note it down. Now we can start the trials:
• Turn the switch on
• Note down the current shown by the ammeter reading
• Allow the current to flow through the circuit for three or four minutes
• Turn off the switch and immediately measure the temperature of water in each beaker.
♦ Let them be tA and tB
• We will see that tA is larger than tB
3. So the observations are over. Let us analyse them:
• Beakers A and B are connected in series. So the same current (indicated by the ammeter) will be flowing through both nichrome and copper
• Both nichrome and copper pieces are of the same size. Because they have the same length and diameter
• Even when current, lengths and diameters are the same, nichrome produced greater heat. Why is that so?
Ans: Nichrome offers greater resistance to the flow of electricity. That is., electricity cannot flow easily through nichrome. So most of the electrical energy will be converted into heat energy.
4. But the opposite happens in the case of copper
• Copper offers lesser resistance to the flow of electricity. That is., electricity can flow easily through copper. So only a lesser electrical energy will be converted into heat energy.
• Since in this trial, current, lengths and diameters of both nichrome and copper are the same, we can say that, resistance to flow of electricity depends on the material with which the conductor is made. We discussed it in a previous chapter here.
■ So we can write: Greater the resistance (R), greater is the heat energy produced
5. Trial 2:
• Allow the water in the beakers to cool down to room temperature.
• Remove the beaker B from the circuit. Now there is only the nichrome conductor in beaker A in the circuit. This is shown in fig.10.1(b) above
• Measure the temperature of water in beaker A and note it down.• Turn on the switch and pass current for five minutes
• Note down the current. Let it be I1
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t1
6. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for seven minutes
• The current must be the same I1 as before. Ensure this by the ammeter reading
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t2
• We will see that, t2 is greater than t1
■ So we can write: Greater the 'time duration (t) for which current is passed through a conductor', greater is the heat energy produced
7. Trial 3:
• Allow the water to cool down to room temperature. Measure this temperature and note it down
• Turn the switch on and allow the current to pass for three minutes
• Note down the current I1
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t1
8. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for the same three minutes
• The current I1 this time must be greater than I1.
♦ Ensure this by the ammeter reading. The current can be increased by adjusting the rheostat
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t2
We will see that t2 is greater than t1
■ So we can write: Greater the 'current (I) which is passed through a conductor', greater is the heat energy produced
From the above activity, we can write:
The heat energy produced in a conductor due to the flow of current through it, depends on three items:
(i) The resistance (R) of the conductor
(ii) The time duration (t) for which the current is passed through the conductor
(iii) The intensity of the current (I)
• The English scientist James Prescott Joule discovered the relation connecting those three factors. The relation is known as Joule's Law:
■ The heat generated in a current carrying conductor is the product of the square of the current (I) in the conductor, the resistance (R) of the conductor and the time (t) of the flow of current
In simple terms, we can say:
The heat will be equal to the product of three items:
(i) Square of the current (I2)
(ii) Resistance (R)
(iii) time (t)
• So if H is the heat generated, then:
Eq.10.1: H = I2Rt
• I is measured in ampere, R in ohm and t in seconds
Let us see some interesting cases:
■ Heat of H1 joules is generated in a conductor when a current I1 passes through it for t seconds. How much heat will be generated if the current is doubled and is passed for the same t seconds?Solution:
• We have: H = I2Rt
• Case 1: H1 = (I1)2Rt
• Case 2: Let H2 be the heat generated when a current I2 passes for the same t seconds
• Then we can write:
H2 = (I2)2Rt
• Taking ratios, we get: H1⁄H2= (I1)2Rt⁄(I2)2Rt = (I1)2⁄(I2)2 = [(I1)⁄(I2)]2
But given that I2 = 2I1
• So we get: H1⁄H2 = [(I1)⁄(2I1)]2 = [(1)⁄(2)]2 = [1⁄4]
⟹ H1⁄H2 = 1⁄4 ⟹ H2 = 4H1
• So we can write: The new heat will be four times the original heat
■ Heat of H1 joules is generated in a conductor when a current I1 passes through it for t seconds. How much heat will be generated if the current is halved and is passed for the same t seconds?
Solution:
• We have: H = I2Rt
• Case 1: H1 = (I1)2Rt
• Case 2: Let H2 be the heat generated when a current I2 passes for t seconds
• Then we can write:
H2 = (I2)2Rt
• Taking ratios, we get: H1⁄H2= (I1)2Rt⁄(I2)2Rt = (I1)2⁄(I2)2 = [(I1)⁄(I2)]2
But given that I2 = 0.5I1
• So we get: H1⁄H2 = [(I1)⁄(0.5I1)]2 = [(1)⁄(0.5)]2 = [(1)⁄(5/10)]2 = [(10)⁄(5)]2 = [2]2 = 4
⟹ H1⁄H2 = 4 ⟹ H2 = H1⁄4
• So we can write: The new heat will be one fourth of the original heat
• The equation H = I2Rt gives us the relation between heat and the quantities: [current, resistance, time]
• Can we relate heat to voltage? Let us try:
• From ohm's law, we have: V = IR (Details here)
• From this we get: I = V⁄R
• So we can use (V⁄R) instead of I
• Thus we get: H = (V⁄R)2Rt ⟹ H = (V2⁄R2)Rt = (V2⁄R)t = V2t⁄R
So we can write:
Eq.10.2: H = V2t⁄R
Another derivation:
• From ohm's law, we have: V = IR
• From this we get: R = V⁄I
• So we can use (V⁄I) instead of R
• Thus we get: H = I2(V⁄I)t ⟹ H = VIt
So we can write:
Eq.10.3: H = VIt
Now we will see a solved example:
Solved example 10.1:
A bulb of resistance 920 Ω works on 230 V supply. Calculate the quantity of heat generated in 3 minutes
Solution:
1. Given: R = 920 Ω, V = 230 volts, t = 3 minutes = (3 × 60) = 180 seconds
2. We want a relation that will connect the heat (H) to: [R, V and t]
• So we will use Eq.10.2: H = V2t⁄R
3. Substituting the values, we get: H = (230)2×180⁄920 = 10350 J
Another method:
1. We have the basic equation: H = I2Rt
2. But the current I is not given. We can calculate it using the Ohm's law: V = IR
• So we get: I = V⁄R = 230⁄920 = 1⁄4 = 0.25 A
3. Thus we get: H = I2Rt = H = (0.25)2 × 920 × 180 = 10350 J
One more method:
1. Here we will try to use Eq.10.3: H = VIt
• We have calculated I in the previous method. So we can use it
2. Substituting the values, we get: H = 230 × 0.25 × 180 = 10350 J
Solved example 10.2
An electric iron works on 230 V. A current of 3 A flows through it for half an hour. Calculate the amount of heat energy generated
Solution:
1. Given: V = 230 volts, I = 3 A, t = 30 minutes = (30 × 60) = 1800 seconds
2. We want a relation that will connect the heat (H) to: [V, I and t]
• So we will use Eq.10.3: H = VIt
3. Substituting the values, we get: H = 230 × 3 × 1800 = 1242000 J = 1242 kJ
■ So we have seen a special property of electricity:
• It will produce heat when it passes through a conductor.
• We also saw how to calculate the 'quantity of heat' produced.
• In the next section we will see some practical applications of this property.
• It will produce heat when it passes through a conductor.
• We also saw how to calculate the 'quantity of heat' produced.
• In the next section we will see some practical applications of this property.
Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved
No comments:
Post a Comment