In the previous section we saw the details about resistivity. In this section, we will see how resistors are connected with each other.
1. Consider the circuit in fig.8.21(a) below:
• It contains: A switch, 2.2 Ω resistor, 1 Ω resistor and 6 V battery.
■ In this circuit, the resistors are connected in series
2. Let us turn on the switch. It is shown in fig.b
• A current 'I' will flow out from the positive terminal of the battery
3. When I passes through the 2.2 Ω resistor, there will be a voltage drop.
• This is because, some energy will be lost as heat in that 2.2 Ω resistor.
4. So the voltage available across the 1 Ω resistor will be low.
• That means, voltage across the two resistors will not be the same.
• Since the voltages are different, we will attach a voltmeter to each resistor.
• They are shown as V1 and V2 in fig.8.22 below:
5. Now consider the fig.8.21(b) again. The current I has no oppurtunity to branch.
• So all the I has to pass through both the resistors. That means, the current through both the resistors will be the same.
• However, to confirm it, we will connect an ammeter to each of the resistors.
• These are shown as A1 and A1 in fig.8.22 above.
6. Also an ammeter A is connected just near the battery. This will give the current flowing out of the battery before it enters any of the resistors.
• The energy supplied by the 6 V battery is used partly by the 2.2 Ω resistor and the rest by the 1 Ω resistor.
• So the total energy used by the resistors should be equal to 6 V.
• To confirm this, we will connect a voltmeter across both the resistors. This is shown as V in the fig.8.22
7. Now, can we guess the readings in the voltmeters and ammeters?
• We can indeed do mathematical calculations and find the answers.
• We will then compare those answers with the readings obtained in the actual experiment.
8. Consider the current I released from the battery.
• How much current will be released?
Ans: It will depend on the resistance that is experienced by the battery
• If the circuit put up a greater 'effective resistance', less current will be released
• If the circuit put up a lesser 'effective resistance', more current will be released
9. There are two resistors. R1 = 2.2 Ω and R2 = 1 Ω
• So what is the 'effective resistance' experienced by the battery?
Ans: We have to find the 'effective resistance' created due to the combined actions of R1 and R2
10. Let this effective resistance be R
• Then, using the relation V = IR, we get: 6 = I×R
11. We know that, this same I will be flowing through R1
♦ So voltage across R1 = V1 = IR1
• We know that,this same I will be flowing through R2
♦ So voltage across R2 = V2 = IR2
12. But (V1+V2) = (Energy in R1 + Energy in R2)
• This sum must be equal to the total energy available from the battery
• So we can write: (V1+V2) = 6 V
13. Substituting for V1 and V2 from (11), we get:
(IR1+ IR2) = 6 V
14. But from (10), we have: 6 V = IR
• So (13) becomes:
(IR1+ IR2) = IR
⟹ I(R1+ R2) = IR
⟹ (R1+ R2) = R
■ This is a very useful result. We can write it as:
When resistors are connected in series, the effective resistance is equal to the sum of the 'individual resistances'.
• If there are n resistors connected in series, we can write: R = (R1+ R2 + . . . + Rn)
• If all those resistors have the same resistance value, we can denote each as r.
♦ Then the effective resistance R = (r + r + . . . + r) [n terms]
♦ Thus we get R = nr
■ So when resistors are connected in series, the effective resistance increases
• For our present case, we get: R = (2.2+1) = 3.2 Ω
15. Once we obtain the effective resistance, we can calculate the current I
• We get: I = V⁄R = 6⁄3.2 = 1.875 A
• Since there is no branching,
♦ Current through R1 = I1 = I = 1.875 A
♦ Current through R2 = I2 = I = 1.875 A
• This is indeed the value obtained from the actual experiment.
♦ See '1.9' written below I1, I2 and I in the table 8.4 below:
16. Now we can calculate the individual voltages:
• The voltage across the 2.2 Ω resistor:
♦ We can use the relation: V = IR
♦ So V1 = IR1 = 1.875 × 2.2 = 4.125 V
[Note that, the reading given by the voltmeter V1 is entered as 4.1 in the table 8.4 above. So the theoretical and experimental results are the same]
• The voltage across the 1 Ω resistor:
♦ We can use the relation: V = IR
♦ So V2 = IR2 = 1.875 × 1 = 1.875 V
[Note that, the reading given by the voltmeter V2 is entered as 1.9 in the table 8.4 above. So the theoretical and experimental results are the same]
17. Let us take the sum (V1+V2):
(4.125 + 1.875) = 6 V
[Note that, the reading given by the voltmeter V is entered as 6 in the table 8.4 above. So the theoretical and experimental results are the same]
1. Consider the circuit in fig.8.21(a) below:
 |
| Fig.8.21 |
■ In this circuit, the resistors are connected in series
2. Let us turn on the switch. It is shown in fig.b
• A current 'I' will flow out from the positive terminal of the battery
3. When I passes through the 2.2 Ω resistor, there will be a voltage drop.
• This is because, some energy will be lost as heat in that 2.2 Ω resistor.
4. So the voltage available across the 1 Ω resistor will be low.
• That means, voltage across the two resistors will not be the same.
• Since the voltages are different, we will attach a voltmeter to each resistor.
• They are shown as V1 and V2 in fig.8.22 below:
 |
| Fig.8.22 |
• So all the I has to pass through both the resistors. That means, the current through both the resistors will be the same.
• However, to confirm it, we will connect an ammeter to each of the resistors.
• These are shown as A1 and A1 in fig.8.22 above.
6. Also an ammeter A is connected just near the battery. This will give the current flowing out of the battery before it enters any of the resistors.
• The energy supplied by the 6 V battery is used partly by the 2.2 Ω resistor and the rest by the 1 Ω resistor.
• So the total energy used by the resistors should be equal to 6 V.
• To confirm this, we will connect a voltmeter across both the resistors. This is shown as V in the fig.8.22
7. Now, can we guess the readings in the voltmeters and ammeters?
• We can indeed do mathematical calculations and find the answers.
• We will then compare those answers with the readings obtained in the actual experiment.
8. Consider the current I released from the battery.
• How much current will be released?
Ans: It will depend on the resistance that is experienced by the battery
• If the circuit put up a greater 'effective resistance', less current will be released
• If the circuit put up a lesser 'effective resistance', more current will be released
9. There are two resistors. R1 = 2.2 Ω and R2 = 1 Ω
• So what is the 'effective resistance' experienced by the battery?
Ans: We have to find the 'effective resistance' created due to the combined actions of R1 and R2
10. Let this effective resistance be R
• Then, using the relation V = IR, we get: 6 = I×R
11. We know that, this same I will be flowing through R1
♦ So voltage across R1 = V1 = IR1
• We know that,this same I will be flowing through R2
♦ So voltage across R2 = V2 = IR2
12. But (V1+V2) = (Energy in R1 + Energy in R2)
• This sum must be equal to the total energy available from the battery
• So we can write: (V1+V2) = 6 V
13. Substituting for V1 and V2 from (11), we get:
(IR1+ IR2) = 6 V
14. But from (10), we have: 6 V = IR
• So (13) becomes:
(IR1+ IR2) = IR
⟹ I(R1+ R2) = IR
⟹ (R1+ R2) = R
■ This is a very useful result. We can write it as:
When resistors are connected in series, the effective resistance is equal to the sum of the 'individual resistances'.
• If there are n resistors connected in series, we can write: R = (R1+ R2 + . . . + Rn)
• If all those resistors have the same resistance value, we can denote each as r.
♦ Then the effective resistance R = (r + r + . . . + r) [n terms]
♦ Thus we get R = nr
■ So when resistors are connected in series, the effective resistance increases
• For our present case, we get: R = (2.2+1) = 3.2 Ω
15. Once we obtain the effective resistance, we can calculate the current I
• We get: I = V⁄R = 6⁄3.2 = 1.875 A
• Since there is no branching,
♦ Current through R1 = I1 = I = 1.875 A
♦ Current through R2 = I2 = I = 1.875 A
• This is indeed the value obtained from the actual experiment.
♦ See '1.9' written below I1, I2 and I in the table 8.4 below:
 |
| Table.8.4 |
• The voltage across the 2.2 Ω resistor:
♦ We can use the relation: V = IR
♦ So V1 = IR1 = 1.875 × 2.2 = 4.125 V
[Note that, the reading given by the voltmeter V1 is entered as 4.1 in the table 8.4 above. So the theoretical and experimental results are the same]
• The voltage across the 1 Ω resistor:
♦ We can use the relation: V = IR
♦ So V2 = IR2 = 1.875 × 1 = 1.875 V
[Note that, the reading given by the voltmeter V2 is entered as 1.9 in the table 8.4 above. So the theoretical and experimental results are the same]
17. Let us take the sum (V1+V2):
(4.125 + 1.875) = 6 V
[Note that, the reading given by the voltmeter V is entered as 6 in the table 8.4 above. So the theoretical and experimental results are the same]
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