Question

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ?​

Answer 1

Given:-

• Voltage Suply(V)=220V
• maximum allowable current =5A
• electric bulb Rated(P)= 10W

To Find:-

• How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A.

Solution:-

Let $R_{1}$ is the resistance of the bulb

We know that

The power dissipated in a resistor is given by :-

$\sf \: P = \frac{{V}^{2}}{R}$

So, we can also write it as

$R _{1}= \frac{{V}^{2}}{ P}$

substituting the value in this equation

$R _{1}= \frac{{220}^{2}}{ 10}$

$= > R _{1}= \frac{220 \times 220}{ 10} Ω$

$= > R _{1}= ( 220 \times 22 )Ω$

$= > R _{1}= 4840Ω$

Now, According to the Ohm's law :

In a Circuit Potential Difference is directly proportional to the current

i.e

$\sf \: V= I \times R$

Let R be the total resistance of the circuit for n number of electric bulbs

now,

$R = \frac{V}{I}$

$= > \frac{220}{5} = 44Ω$

Resistance for each electric bulb is:-

$R_1=4840Ω$

and we know that lamps are connected in parallel

$\sf \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} ..n \: times$

So,

$= > \frac{1}{R} = \frac{n}{R_1}$

$= > n = \frac{R_1}{R}$

$= > n = \frac{4840}{44}$

$= > n = 110$

∴ There are 110 lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A.