RESISTORS IN SERIES
When several resistors are joined in series,
the resistance of the combination is always
equal to the sum of their individual
Effective resistance is always greater than
any individual resistance
It is observed that the total current I, is equal to
the sum of the separate currents through each
branch of the combination.
Let RP be the equivalent resistance of the
parallel combination of resistors.
By applying Ohm’s law to the parallel
combination of resistors. We have,
I=V/Rp ; 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 we may conclude that 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
Effective resistance is always lesser than
any individual resistance when connected in
Current is Constant Voltage is Constant
Voltage varies across
Current flowing through the
resistors varies with different
Effective resistance will be
greater than any individual
RS = R1+ R2 + R3…….Rn
Effective resistance will be
less than any least individual
𝑅𝑝 = 1
ADVANTAGE OF PARALLEL ELECTRIC
• when you turn ON one gadget ,you don’t necessarily want to turn on all the
• Individual switches can be provided to each device.
Allows for additional components:
• Parallel circuits allow components to be added in the circuit without changing
• For example if you need an additional light in your home you can add one new
bulb, it doesn't affect the current through other bulbs.
•If “n” resistors of equal resistances are connected
in series then the effective resistance would be nR.
•If “n” resistors of equal resistance are connected in
parallel then the effective resistance would be R/n.
Q, If n equal resistors are connected in series and
then connected in parallel, the ratio of the
maximum to minimum resistance is__________
(a) n (b) 1 / n2 (c) n2 (d) 1 / n
SOLUTION FOR PROBLEM 1
When n resistors are connected in series,
Equivalent resistance Rn= nR (Maximum)
When connected in parallel,
Equivalent resistance Rp=R/n (minimum)
The ratio of the maximum to minimum resistance is,
nR /(R/n) = n2
Q, Three resistances each of 4 ohm are
connected to form a triangle. The
resistance between any two terminal is
If we switch ON the circuit , then the bulb will start to
glow. In order to keep bulb glowing battery have to
supply current continuously by expending its energy.
Does all the energy spent by the source (battery) is
utilized by the bulb?
Yes or No?
ANSWER IS “NO”
• Because as the bulb started to glow, it consumes
electrical energy and convert it into light energy but not
• Some of the energy is dissipated as heat in and around
the bulb.( you may have the experience while you touch
an incandescent lamp which is kept ON for few minutes
you feel it hot)
• Another example : Electric fan becomes warm if it is used
continuously for long time.
APPLICATIONS OF THIS
• The energy wasted (in the form of heat) is an inevitable
consequence of current.
• Even though it is undesirable and unavoidable it gives
insight that this heating effect can be utilized in some
devices..(ex: Electric heater, electric iron etc..)
• If the electric circuit is purely resistive then the entre
energy from the source is converted in the form of HEAT.
JOULES LAW OF HEATING
current I flowing
through a resistor
of resistance R.
Let the potential
it be V.
• When electric current flows through resistors (which offers
resistance) suffer obstruction.
• Work has to be done to overcome this resistance , which is
converted into heat energy.
• If Q amount of charge flows through potential difference V, then
• W= Q x V since [ W/Q = V]
• W = Q x V ------------- (1)
• I = Q / t
• Q = I x t -------------- (2)
From (1) & (2)
W = V I t -------------- (3)
By ohms law , V = I x R
W = (IR) It = I 2 R t
This work is converted into energy
H = I 2 R t ---------------- (4)
The above equation is called joules law of heating.
PROBLEM - 4
•A current of 2A passing through conductor
produces 80 J of heat in 10 seconds. The resistance
of the conductor is
(a) 0.5 Ω (b) 2 Ω (c) 4 Ω (d) 20 Ω
•Two heater wires of same dimension are first
connected in series and then in parallel to a source
of power supply. What will be the ratio of heat
produced in the two cases ?
SOLUTION FOR PROBLEM 5
As heater wires are of the same dimensions, their
resistance are equal, say R each.
In series, their resultant resistance Rs=2R.
Now, if connected to supply source of voltage V for
time t, the heat produced,
Hs= V2t/Rs = V2t/2R
In parallel, the resultant resistance,
And for same supply voltage V and
for the same time t, the heat
Hp= V2t/Rp = 2V2t/R
• It protects circuits and appliances by stopping the
flow of any excessive high electric current.
• It consists of a metal wire (ex: copper,
aluminium….etc.) of low melting point.
• It is placed always in series before the device.
• If a current is larger than the specified value flows
through the circuit , the temperature of the fuse
wire increases then melts the fuse wire and breaks
• Fuse wire of various current capacities
(1A,2A,5A…etc )is used for domestic purpose.
• For an electric iron which consumes 1kW power
when operates at 220V. In order to protect the
device choose a suitable fuse.
P = V x I or I =
= 4.5 A
So 5A fuse must be used to protect the circuit.