# 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 *R _{1}*,

*R*&

_{2}*R*(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.

_{3}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
*V _{1}*,

*V*and

_{2}*V*across the first, second and third resistors separately.

_{3 }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 = V _{1}* +

*V*+

_{2}*V*

_{3}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,

*V _{1 }=I
R_{1}, V_{2 }=I
R_{2 }V_{3
}=I R_{3}.*

*I R = I R _{1}* +

*I R*+

_{2}*I R*or

_{3 }*R*

_{s }= R_{1}+R_{2}+ R_{3}When
several resistors are joined in series, resistance of the combination *R _{s}* equals the sum of their
individual resistances,

*R*.

_{1}, R_{2}, R_{3}**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, *R _{1} *= 20 Ω,

Resistance of the conductor connected in series, *R _{2}* = 4 Ω.

Then
the total resistance, *R = R _{1} +
R_{2}*

*R _{s
}= *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*/*R*_{s}

= 6 V/24 Ω
= __0.25 A.__

c) Potential
difference across the electric lamp:

*V _{1 }*=
20 Ω × 0.25 A =

__5 V__

Potential
difference across the conductor:

*V _{2}*
= 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 R_{1},
R_{2}, & R_{3} 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 *R _{1}*, as shown below. Note the ammeter reading,

*I*Similarly, measure the currents

_{1}.*I*& I

_{2}_{3 }through

*R*&

_{2}*R*respectively.

_{3}*I = I _{1} + I_{2} + I_{3}*

Let *R _{p}* be the equivalent resistance of the parallel
combination of resistors.

Hence, *I = V/R _{p }*

On applying Ohm’s law to each
resistor, we have

*I _{1} = V /R_{1}*

*I*

_{2}= V /R_{2}*I*

_{3}= V /R_{3}*V/R _{p} = V/R_{1} + V/R_{2} + V/R_{3}*

or

1*/R _{p} = *1

*/R*1

_{1}+*/R*1

_{2}+*/R*

_{3 }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 *R _{1}*,

*R*&

_{2}*R*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.

_{3}

**Solution:**

*R _{1}*
= 5 Ω,

*R*= 10 Ω, and

_{2}*R*= 30 Ω.

_{3}Potential difference across the
battery, *V* = 12 V.

This is also the potential
difference across each of the individual resistor.

The
current *I _{1}*, through

*R*

_{1}= V/ R_{1}*I _{1 }= *12
V/5 Ω = 2.4
A.

The
current *I _{2}*, through

*R*

_{2 }= V/ R_{2}*I _{2 }= *12

*V/10 Ω = 1.2 A.*

The
current *I _{3}*, through

*R*

_{3 }= V/R_{3}*I _{3 }= *12

*V/30 Ω = 0.4 A.*

The total current in the circuit,
I = I_{1} + I_{2} + I_{3}

= (2.4 + 1.2 + 0.4) A = 4 A

The total resistance *R _{p}*, is 1

*/R*1

_{p}=*/R*1

_{1}+*/R*1

_{2}+*/R*

_{3}

Thus, *R _{p}* = 3 Ω.

**Problem: **If
in Fig. given below, *R _{1}* =
10 Ω,

*R*40 Ω,

_{2}=*R*30 Ω,

_{3}=*R*20 Ω,

_{4}=*R*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.

_{5}=

**Solution:**

Suppose we replace the parallel
resistors *R _{1}* and

*R*by an equivalent resistor of resistance,

_{2}*R*′. Similarly, we replace the parallel resistors

*R*,

_{3}*R*and

_{4}*R*by an equivalent single resistor of resistance

_{5}*R*″.

1*/R _{p} = *1

*/R*1

_{1}+*/R*1

_{2}+*/R*

_{3}∴ 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.

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