In the previous section we saw an example of 'resistors connected in series'. In this section, we will see an example of 'resistors connected in parallel'.
1. Consider the circuit in fig.8.23(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 parallel
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 reaches the junction X, it will split into I1 and I2
• When I1 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. When I2 passes through the 1 Ω resistor, there will be a voltage drop.
• This is because, some energy will be lost as heat in that 1 Ω resistor.
• we will attach a voltmeter to each resistor.
• They are shown as V1 and V2 in fig.8.24 below:
5. Now consider the fig.8.23(b) again. The current I has split into I1 and I2
• To measure I1 and I2 separately, we will connect an ammeter to each of the resistors.
• These are shown as A1 and A1 in fig.8.24 above.
• Also note that, at junction Y, currents I1 and I2 will combine together to form I again
6. 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.
• Also, to measure the total voltage, we will connect a voltmeter across both the resistors. This is shown as V in the fig.8.24
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. When I1 flows through R1, there will be a voltage drop.
• But this does not affect the R2. Because R2 is placed independently from R1
• So both R1 and R2 will get the same 6 V energy from the battery. We can write:
♦ Voltage (V1) across R1 = V = 6 = I1 × R1
♦ Voltage (V2) across R2 = V = 6 = I2 × R2
• This is indeed the value obtained from the actual experiment.
♦ See '6' written below V1, V2 and V in the table 8.5 below.
12. From the above two equations, we get:
• I1 = V⁄R1
• I2 = V⁄R2
• So (I1 + I2) = (V⁄R1 + V⁄R2)
13. But (I1 + I2) = I
• So the result in (12) becomes:
I = (V⁄R1 + V⁄R2)
14. From (10), we have: V = IR. Where R is the effective resistance in the circuit
• We can rearrange this equation and write: I = V⁄R
• Substituting this in (13), we get:
V⁄R = (V⁄R1 + V⁄R2)
⟹ 1⁄R = (1⁄R1 + 1⁄R2)
■ This is a very useful result. We can write it as:
(i) When two resistors R1 and R2 are connected in parallel, the effective resistance R is given by:
1⁄R = (1⁄R1 + 1⁄R2)
(ii) When three resistors R1, R2 and R3 are connected in parallel, the effective resistance R is given by:
1⁄R = (1⁄R1 + 1⁄R2+ 1⁄R3)
(iii) When n resistors R1, R2, R3 . . . Rn are connected in parallel, the effective resistance R is given by:
1⁄R = (1⁄R1 + 1⁄R2+ 1⁄R3 + . . . + 1⁄Rn)
(iv) When n resistors R1, R2, R3 . . . Rn are connected in parallel, and they have the same resistance, we can denote each of them by r. Then the effective resistance R is given by:
1⁄R = (1⁄r + 1⁄r + 1⁄r + . . . + 1⁄r) [n terms]
So we get: 1⁄R = (n⁄r)
⟹ R = r⁄n.
The above formula in (iv) gives very interesting results. Let us analyse it:
■ Let the number of resistors be 2. Then n = 2
• Remember that both the resistors should have the same resistance of r Ω
♦ Then only this formula will work
• Then the effective resistance 'R' that the battery will experience is r⁄2
• An example: Let each of the two resistors in parallel be of 10 Ω.
♦ Then the battery will experience an effective resistance of only (10⁄2) = 5 Ω
■ Let the number of resistors be 3. Then n = 3
• Remember that all the three resistors should have the same resistance of r Ω
♦ Then only this formula will work
• Then the effective resistance 'R' that the battery will experience is r⁄3
• An example: Let each of the three resistors in parallel be of 12 Ω
♦ Then the battery will experience an effective resistance of only (12⁄3) = 4 Ω
■ Let the number of resistors be 4. Then n = 4
• Remember that all the three resistors should have the same resistance of r Ω
♦ Then only this formula will work
• Then the effective resistance 'R' that the battery will experience is r⁄4
• An example: Let each of the four resistors in parallel be of 12 Ω
♦ Then the battery will experience an effective resistance of only (12⁄4) = 3 Ω
• Note that, when 3 nos. of 12 Ω resistors are used in parallel, R = 4 Ω
• When 4 nos. of 12 Ω resistors are used in parallel, R = 3 Ω
■ So adding more resistors in parallel has the effect of decreasing the effective resistance R
• This is because, when a resistor is added, a new path is being opened up for the flow of current.
• Thus the resistance to the flow of current is being reduced.
• As a consequence, the current I in the circuit will increase
• For our present case, we have: 1⁄R = (1⁄R1 + 1⁄R2)
⟹ 1⁄R = (1⁄2.2 + 1⁄1) = 1.4545
⟹ R = 1⁄1.4545 = 0.6875 Ω
15. Once we obtain the effective resistance, we can calculate the current I
• We get: I = V⁄R = 6⁄0.6875 = 8.7272 A
• This is indeed the value obtained from the actual experiment.
♦ See '8.7' written below A in the table 8.5 below:
• Since there is branching, we need to find I1 and I2 separately:
♦ Current through R1 = I1 = V⁄R1 = 6⁄2.2 = 2.7272 A
♦ Current through R2 = I2 = V⁄R2 = 6⁄1 = 6 A
♦ The sum (I1 + I2) = (2.72 + 6) = 8.72 A
• These are indeed the values obtained from the actual experiment.
♦ See '2.7' written below A1, '6' written below A2 and '8.7' written below A in the table 8.5 below:
1. Consider the circuit in fig.8.23(a) below:
 |
| Fig.8.23 |
■ In this circuit, the resistors are connected in parallel
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 reaches the junction X, it will split into I1 and I2
• When I1 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. When I2 passes through the 1 Ω resistor, there will be a voltage drop.
• This is because, some energy will be lost as heat in that 1 Ω resistor.
• we will attach a voltmeter to each resistor.
• They are shown as V1 and V2 in fig.8.24 below:
 |
| Fig.8.24 |
• To measure I1 and I2 separately, we will connect an ammeter to each of the resistors.
• These are shown as A1 and A1 in fig.8.24 above.
• Also note that, at junction Y, currents I1 and I2 will combine together to form I again
6. 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.
• Also, to measure the total voltage, we will connect a voltmeter across both the resistors. This is shown as V in the fig.8.24
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. When I1 flows through R1, there will be a voltage drop.
• But this does not affect the R2. Because R2 is placed independently from R1
• So both R1 and R2 will get the same 6 V energy from the battery. We can write:
♦ Voltage (V1) across R1 = V = 6 = I1 × R1
♦ Voltage (V2) across R2 = V = 6 = I2 × R2
• This is indeed the value obtained from the actual experiment.
♦ See '6' written below V1, V2 and V in the table 8.5 below.
12. From the above two equations, we get:
• I1 = V⁄R1
• I2 = V⁄R2
• So (I1 + I2) = (V⁄R1 + V⁄R2)
13. But (I1 + I2) = I
• So the result in (12) becomes:
I = (V⁄R1 + V⁄R2)
14. From (10), we have: V = IR. Where R is the effective resistance in the circuit
• We can rearrange this equation and write: I = V⁄R
• Substituting this in (13), we get:
V⁄R = (V⁄R1 + V⁄R2)
⟹ 1⁄R = (1⁄R1 + 1⁄R2)
■ This is a very useful result. We can write it as:
(i) When two resistors R1 and R2 are connected in parallel, the effective resistance R is given by:
1⁄R = (1⁄R1 + 1⁄R2)
(ii) When three resistors R1, R2 and R3 are connected in parallel, the effective resistance R is given by:
1⁄R = (1⁄R1 + 1⁄R2+ 1⁄R3)
(iii) When n resistors R1, R2, R3 . . . Rn are connected in parallel, the effective resistance R is given by:
1⁄R = (1⁄R1 + 1⁄R2+ 1⁄R3 + . . . + 1⁄Rn)
(iv) When n resistors R1, R2, R3 . . . Rn are connected in parallel, and they have the same resistance, we can denote each of them by r. Then the effective resistance R is given by:
1⁄R = (1⁄r + 1⁄r + 1⁄r + . . . + 1⁄r) [n terms]
So we get: 1⁄R = (n⁄r)
⟹ R = r⁄n.
The above formula in (iv) gives very interesting results. Let us analyse it:
■ Let the number of resistors be 2. Then n = 2
• Remember that both the resistors should have the same resistance of r Ω
♦ Then only this formula will work
• Then the effective resistance 'R' that the battery will experience is r⁄2
• An example: Let each of the two resistors in parallel be of 10 Ω.
♦ Then the battery will experience an effective resistance of only (10⁄2) = 5 Ω
■ Let the number of resistors be 3. Then n = 3
• Remember that all the three resistors should have the same resistance of r Ω
♦ Then only this formula will work
• Then the effective resistance 'R' that the battery will experience is r⁄3
• An example: Let each of the three resistors in parallel be of 12 Ω
♦ Then the battery will experience an effective resistance of only (12⁄3) = 4 Ω
■ Let the number of resistors be 4. Then n = 4
• Remember that all the three resistors should have the same resistance of r Ω
♦ Then only this formula will work
• Then the effective resistance 'R' that the battery will experience is r⁄4
• An example: Let each of the four resistors in parallel be of 12 Ω
♦ Then the battery will experience an effective resistance of only (12⁄4) = 3 Ω
• Note that, when 3 nos. of 12 Ω resistors are used in parallel, R = 4 Ω
• When 4 nos. of 12 Ω resistors are used in parallel, R = 3 Ω
■ So adding more resistors in parallel has the effect of decreasing the effective resistance R
• This is because, when a resistor is added, a new path is being opened up for the flow of current.
• Thus the resistance to the flow of current is being reduced.
• As a consequence, the current I in the circuit will increase
• For our present case, we have: 1⁄R = (1⁄R1 + 1⁄R2)
⟹ 1⁄R = (1⁄2.2 + 1⁄1) = 1.4545
⟹ R = 1⁄1.4545 = 0.6875 Ω
15. Once we obtain the effective resistance, we can calculate the current I
• We get: I = V⁄R = 6⁄0.6875 = 8.7272 A
• This is indeed the value obtained from the actual experiment.
♦ See '8.7' written below A in the table 8.5 below:
• Since there is branching, we need to find I1 and I2 separately:
♦ Current through R1 = I1 = V⁄R1 = 6⁄2.2 = 2.7272 A
♦ Current through R2 = I2 = V⁄R2 = 6⁄1 = 6 A
♦ The sum (I1 + I2) = (2.72 + 6) = 8.72 A
• These are indeed the values obtained from the actual experiment.
♦ See '2.7' written below A1, '6' written below A2 and '8.7' written below A in the table 8.5 below:
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| Table.8.5 |
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