ELECTRICITY
RESISTANCE
OF A SYSTEM OF RESISTORS
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.
Resistors in Series
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.
Solution:
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
R=
V/I
= 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
or
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.
Solution:
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.
Solution:
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.
