(ii) Parallel connections  
Resistors in series  
Resistors are said to be in series when two  
or more resistors are connected end – to –  
end consequently  
( )  
V = V1 + V2 + V3 … … . . i  
( )  
From Ohm’s law V = I R … … … ii  
Equation (i) can be written as  
I RT = IR1 + IR2 + IR3  
Fact out “I” both sides  
Properties of series connection  
(i) Current passing through the resistors is  
the same at all points around the circuit.  
I (RT) = I(R1 + R2 + R3)  
Divide by “I” both sides to get,  
RT = R1 + R2 + R3 … … . (iii)  
퐼1 = 퐼2 = 퐼3 = 퐼  
(ii) The voltage across the resistors divides.  
The total p.d is given by  
Therefore, the total resistance in series is the  
sum of the individual resistances in the  
given circuit.  
푉 = 푉 + 푉 + 푉  
1
2
3
(iii) Total (equivalent) resistance is the sum  
of the individual resistance.  
Example 01  
Find the total resistance in the circuit below  
푅푇 = 푅1 + 푅2 + 푅3 + … . . . +푅푛  
(iv) Series connection produces high  
resistance which reduces the current flow.  
Derivation of the formula  
Consider the diagram below for series  
connection  
Solution  
RT = R1 + R2  
RT = 5.0 Ω + 4.0 Ω  
RT = 9.0 Ω  
Example 02  
Three resistorsR1, R2 and R3 are joined in  
series. By using a circuit diagram, find the  
equivalent resistance.  
Solution  
Resistors are said to be in series when two  
or more resistors are connected end – to –  
end consequently  
Step 1: To find the total resistance in the  
circuit  
RT = R1 + R2  
RT = 6 Ω + 4 Ω  
RT = 10 Ω  
Step 2: To find the current in the circuit  
V
I =  
R
20 V  
I =  
( )  
V = V1 + V2 + V3 … … . . i  
10 Ω  
( )  
From Ohm’s law V = I R … … … ii  
Equation (i) can be written as  
I RT = IR1 + IR2 + IR3  
I = 2A  
(b) Find the p.d across the 6 Ω resistor.  
Fact out “I” both sides  
Solution  
I (RT) = I(R1 + R2 + R3)  
Divide by “I” both sides to get,  
RT = R1 + R2 + R3  
V = I R  
V = 2A × 6 Ω  
V = 12 V  
Therefore, the p.d across the 6 Ω resistor =  
12 V  
Example 03  
A resistor of 6 Ω is connected in series with  
a resistor of 4 Ω resistances. A potential  
difference of 20 V is applied across the  
combination.  
Example 04  
The two resistors are connected in series as  
shown in the circuit diagram below.  
(a) Calculate the current through the circuit.  
Solution  
Consider the diagram below  
Example 05  
The figure below shows an electric circuit  
where, 푉퐵 = 9푉 , 푅1 = 4 Ω, 푅2 = 5 Ω and  
푅3 = 6 Ω  
(a) What is the current in the 5 Ω resistor?  
Solution  
V
I =  
R
10 V  
I =  
50 Ω  
I = 2A  
(a) What is the total resistance of the circuit?  
Solution  
(b) What is the current through R?  
Answers  
RT = R1 + R2 + R3  
RT = 4 Ω + 5 Ω + 6 Ω  
RT = 15 Ω  
Since the two resistors are in series, the  
current is the same. The current through R =  
2 A  
(b) What is the value of the current flowing  
in the circuit?  
(c) What is the value of R?  
Solution  
Solution  
V
R =  
I
V
I =  
R
6 V  
I =  
9 V  
I =  
2 A  
15 Ω  
I = 3 Ω  
I = 0.6A  
(c) What is the p.d across each resistor?  
Solution  
(d) What is the value of the p.d , V?  
Solution  
Case I: P.d across 푹ퟏ  
V1 = IR1  
V = V1 + V2  
V = 10 V + 6 V  
V = 16 V  
V1 = 0.6 A × 4 Ω  
V1 = 2.4 V  
The brightness of the bulbs decreases when  
bulbs are connected in series.  
This is because, the total resistance increases  
when more bulbs (resistors) are added in  
series and consequently the potential  
difference across the individual bulbs is  
reduced.  
Case II: P.d across 푹ퟐ  
V2 = IR2  
V2 = 0.6 A × 5 Ω  
Disadvantages of connecting bulbs in  
series  
V2 = 3.0 V  
Case III: P.d across 푹ퟑ  
V3 = IR3  
V3 = 0.6 A × 6 Ω  
V3 = 3.6 V  
1. If one bulb fails, all go off  
In series circuit, the same path is shared. If  
one bulb burns out or is removed, the circuit  
breaks and every bulb stop working.  
In the circuit below, if bulb A blows off,  
other bulbs B and C stops working because,  
the current path is broken. If all switches are  
closed, Bulb A will light brighter since all  
current in the circuit flows through it.  
(d) What is the electric potential at point A?  
Solution  
This can be calculated by taking the p.d  
between two points starting from the  
negative terminal as point of reference  
where the p.d = 0V, the points are 푅3 and  
푅2  
VA = VR + VR  
3
2
VA = 3.6 V + 3.0 V  
VA = 6.6 V  
Effects of connecting bulbs in series  
2. Bulbs becomes dimmer  
Consider the diagram below which shows  
two bulbs connected in series  
Because the bulbs share the same voltage,  
each bulb gets only a fraction of the total  
voltage. This makes the bulbs glow dimmer  
compared to when connected in parallel.  
3. No individual control  
You cannot switch one bulb on or off  
independently. Turning off one turns off all  
unless special switches are used.  
4. Voltage drop issues