In electrical gadgets, resistors are used in various combinations based on Ohm’s law.
There are 2 methods of joining the resistors: Resistors in series & Resistors in parallel.
Join three resistors having resistances R1, R2 & R3 (e.g. 1 Ω, 2 Ω, 3 Ω) in series. Connect them with a 6 V battery, an ammeter and a plug key. Note the ammeter reading.
Change the position of ammeter to anywhere in between the resistors.
The value of the current in the ammeter is the same. i.e., in a series combination of resistors, the current is the same in every part of the circuit or the same current through each resistor.
Insert a voltmeter across the ends X and Y of the series combination of three resistors. Note its reading. It gives potential difference (V) across the series combination of resistors. Now measure the potential difference across the two terminals of the battery. Compare the two values.
Now insert the voltmeter across the ends X and P of the first resistor, as shown below.
Measure the potential differences V1, V2 and V3 across the first, second and third resistors separately.
The total potential difference V across a combination of resistors in series is equal to the sum of potential differences across the individual resistors.
i.e., V = V1 + V2 + V3
Let I be the current through this electric circuit. The current through each resistor is also I. The three resistors joined in series can be replaced by an equivalent single resistor of resistance R, such that the potential difference and the current remains the same. Applying the Ohm’s law to the entire circuit, we have
V = I R
On applying Ohm’s law to the three resistors separately,
V1 =I R1, V2 =I R2 V3 =I R3.
I R = I R1 + I R2 + I R3 or Rs = R1 +R2 + R3
When several resistors are joined in series, resistance of the combination Rs equals the sum of their individual resistances, R1, R2, R3.
Problem: An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.
a) Resistance of electric lamp, R1 = 20 Ω,
Resistance of the conductor connected in series, R2 = 4 Ω.
Then the total resistance, R = R1 + R2
Rs = 20 Ω + 4 Ω = 24 Ω.
b) The total potential difference across the two terminals of the battery V = 6 V.
The current through the circuit is I = V/Rs
= 6 V/24 Ω = 0.25 A.
c) Potential difference across the electric lamp:
V1 = 20 Ω × 0.25 A = 5 V
Potential difference across the conductor:
V2 = 4 Ω × 0.25 A = 1 V
If we replace the series combination of electric lamp and conductor by a single and equivalent resistor, its resistance R would be
= 6 V/ 0.25 A = 24 Ω
This is the total resistance of the series circuit; it is equal to the sum of the two resistances.
Resistors in Parallel
Make a parallel combination, XY, of three resistors having resistances R1, R2, & R3 in an electric circuit. Connect a voltmeter in parallel with the resistors.
Note the ammeter reading (I) and the voltmeter reading.
Voltmeter shows the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor.
Insert the ammeter in series with the resistor R1, as shown below. Note the ammeter reading, I1. Similarly, measure the currents I2 & I3 through R2 & R3 respectively.
I = I1 + I2 + I3
Let Rp be the equivalent resistance of the parallel combination of resistors.
Hence, I = V/Rp
On applying Ohm’s law to each resistor, we have
I1 = V /R1 I2 = V /R2 I3 = V /R3
V/Rp = V/R1 + V/R2 + V/R3
1/Rp = 1/R1 + 1/R2 + 1/R3
Thus, the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
Problem: In the circuit diagram given, suppose the resistors R1, R2 & R3 have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) total current in the circuit, and (c) total circuit resistance.
R1 = 5 Ω, R2 = 10 Ω, and R3 = 30 Ω.
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual resistor.
The current I1, through R1 = V/ R1
I1 = 12 V/5 Ω = 2.4 A.
The current I2, through R2 = V/ R2
I2 = 12 V/10 Ω = 1.2 A.
The current I3, through R3 = V/R3
I3 = 12 V/30 Ω = 0.4 A.
The total current in the circuit, I = I1 + I2 + I3
= (2.4 + 1.2 + 0.4) A = 4 A
The total resistance Rp, is 1/Rp = 1/R1 + 1/R2 + 1/R3
Thus, Rp = 3 Ω.
Problem: If in Fig. given below, R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.
Suppose we replace the parallel resistors R1 and R2 by an equivalent resistor of resistance, R′. Similarly, we replace the parallel resistors R3, R4 and R5 by an equivalent single resistor of resistance R″.
1/Rp = 1/R1 + 1/R2 + 1/R3
∴ 1/R′ = 1/10 + 1/40 = 5/40. i.e., R′ = 8 Ω
1/R″ = 1/30 + 1/20 + 1/60 = 6/60. i.e., R″ = 10 Ω
Thus, the total resistance, R = R′ + R″ = 18 Ω
The current in circuit, I = V/R = 12 V/18 Ω = 0.67 A
Disadvantages of Series Circuit:
- In a series circuit, the current is constant throughout the electric circuit. So, it is impracticable to connect an electric bulb and an electric heater in series, because they need currents of different values.
- When one component fails, the circuit is broken and none of the components works. E.g. it is very difficult to locate the dead bulb in fairy lights.
Advantages of Parallel Circuit:
- A parallel circuit divides the current through the electrical gadgets.
- The total resistance in a parallel circuit is decreased. This is helpful when each gadget has different resistance and requires different current to operate properly.