The more bulbs added, the greater the total  
resistance of the circuit. This causes less  
current to flow, reducing the brightness.  
Consider the circuit diagram below  
5. Not suitable for appliances with  
different rating  
Bulbs or device may require different  
voltages. In series, they all get the same  
current but not required voltage division,  
which can damage some devices.  
Resistors in parallel  
Resistors are said to be in parallel  
connection when two or more resistors are  
placed side by side with their corresponding  
ends joined together  
The total current is given by:  
I = I1 + I2 + I3 … … … (i)  
From Ohm’s law  
V
I =  
… … … … . (ii)  
R
Equation (i) can be expressed as  
V
V
V
V
=
+
+
RT  
R1  
R2  
R3  
Fact out “V” both sides we get  
1
1
1
1
V (  
) = V(  
+
+
)
RT  
R1  
R2  
R3  
Properties of parallel connection  
Divide by “V” both sides to get  
(i) The potential difference across each  
resistor is the same as the total voltage in the  
circuit  
1
1
1
1
=
+
+
RT  
R1  
R2  
R3  
V = V1 = V2 = V3  
(ii) The current flowing in each branch is  
different. The total current is given by:  
Example 01  
In the circuit shown in the figure below, the  
battery has a voltage, V = 10V, 푅1 = 4 Ω,  
푅2 = 5 Ω and 푅3 = 6 Ω  
I = I1 + I2 + I3  
(iii) The total (effective) resistance is given  
by:  
1
1
1
1
1
=
+
+
+ … +  
RT  
R1  
R2  
R3  
Rn  
(iv) Parallel connection gives a low  
resistance than the smallest in the circuit.  
Derivation of the formula  
V
I1 =  
R1  
10 V  
=
4 Ω  
= 2.5 A  
Step 2: To find the current through the 5 Ω  
V
I2 =  
R2  
From the circuit, find:  
10 V  
=
5 Ω  
(a) Effective resistance and total current  
flowing  
= 2.0 A  
Solution  
Step 3: To find the current through the 6 Ω  
Step 1: To find the total resistance  
V
I3 =  
R3  
1
1
1
1
=
+
+
RT  
R1  
R2  
R3  
10 V  
=
6 Ω  
1
1
1
1
=
+
+
RT  
4
5
6
= 1.67 A  
1
15 + 12 + 10  
=
RT  
60  
(c) Sum of the current flowing  
Solution  
1
37  
=
RT  
60  
I = I1 + I2 + I3  
I = 2.5A + 2.0A + 1.67A  
I = 6.17A  
60 Ω  
RT =  
37  
RT = 1.62 Ω  
Step 2: To find the current  
Example 02  
V
I =  
R
Two 40Ω resistors and a 20Ω resistor all are  
connected in parallel with a 12 V power  
supply.  
10 V  
I =  
1.62 Ω  
(a) Draw the circuit diagram, to show the  
arrangement and indicate the direction of  
flow current.  
I = 6.17 A  
Answers  
(b) The current flowing through each  
resistor  
Solution  
Step 1: To find the current through the 4 Ω  
I1 = 0.3 A  
Since, I1 = I2, therefore I1 0.3A and  
I2 0.3A  
Step 4: To find the current, I3  
V
I3 =  
R3  
12 V  
I3 =  
20 Ω  
(b) Determine the current flowing in each  
wire  
I3 = 0.6 A  
Example 03  
Solution  
Carefully study the circuit below and use it  
to answer the next questions  
Step 1: To calculate the total resistance in  
the circuit  
1
1
1
1
=
=
+
+
+
+
RT  
R1  
R2  
R3  
1
1
1
1
RT  
40  
40  
20  
1
1 + 1 + 2  
=
RT  
40  
1
4
=
RT  
40  
(a) What is the potential difference across  
each resistor?  
40 Ω  
4
RT =  
Answers  
RT = 10 Ω  
Since they are in parallel, the p.d across each  
resistor is the same as the total voltage  
across the source terminals  
Step 2: To find the total current, I  
V
I =  
R
Therefore, the p.d = 6 V  
12 V  
I =  
10 Ω  
(b) Calculate the total current drawn from  
the cell  
I = 1.2 A  
Solution  
Step 3: To find the current, I1 and I2  
Step 1: To calculate the total resistance in  
the circuit  
V
I1 =  
R1  
1
1
1
12 V  
I1 =  
=
+
RT  
R1  
R2  
40 Ω  
1
1
1
Connecting bulbs in parallel is advisable for  
several practical reasons as follows:  
=
+
RT  
6
12  
1
2 + 1  
=
(i)  
Bulbs get full voltage  
RT  
12  
1
3
In parallel connections, all bulbs  
receive the same voltage as the  
power source. This ensures each  
bulb glows at its full brightness.  
=
RT  
12  
12 Ω  
RT =  
3
RT = 4 Ω  
(ii)  
Independent operations  
Step 2: To find the total current, I  
If one bulb burns out or is  
V
I =  
R
removed,  
the  
others  
keep  
working. This is because each  
bulb has its own separate circuit  
path.  
6 V  
I =  
4 Ω  
I = 1.5 A  
Effects of connecting bulbs in parallel  
Consider two bulbs in parallel as shown on  
the diagram below  
If the lamp 1 burns out, lamp 2 will continue  
working because the current path in lamp 2  
is not broken  
The ammeter 퐴1 measures current through  
lamp 퐿1 while ammeter 퐴2 measures current  
through lamp 퐿2  
The brightness of the individual bulbs  
remains the same with an increase of the  
number of bulbs in the circuit.  
This is because the potential difference  
across each bulb remains the same when the  
number of bulbs is increased in parallel  
connection.  
(iii)  
Consistency brightness  
Adding more bulbs in parallel  
does not affect the brightness of  
NB: It seem that, when connected in  
parallel the bulbs are brighter, this because  
of the combined brightness of all the bulbs  
others,  
unlike  
the  
series  
connection where the voltage  
gets divided.  
(iv)  
Easier to match the devices  
Advantages of parallel connection of  
bulbs