Question

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

Answer 1

Hint
In the given question, we have been provided with the power rating of the bulbs, the voltage across the circuit and the maximum allowable current. We could apply ohm’s law but we need the equivalent resistance of the combination of bulbs; to obtain that we need the value of the resistance of one bulb. We can use the power rating expression to find the resistance of one of the electric bulbs. Let’s see the detailed solution.
$\Rightarrow P=\dfrac{{{V}^{2}}}{R}$ , $V=IR$ , $\Rightarrow \dfrac{\text{1}}{{{\text{R}}_{\text{parallel}}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+--+\dfrac{1}{{{R}_{n}}}$

Complete step by step answer
We have been told that the electric bulbs, each of power rating $(P)=10W$ are designed for a voltage source of $220V$
We know that for an appliance with a given power rating, we have $P=\dfrac{{{V}^{2}}}{R}$ where $R$ is the resistance of the appliance
Substituting the values, we can obtain the resistance of one of the bulbs as follows
$\Rightarrow 10=\dfrac{{{\left( 220 \right)}^{2}}}{R}$
$\Rightarrow R=\dfrac{48400}{10}=4840\Omega $
Now that we have obtained the value of the resistance of one bulb, we should discuss about parallel combination
For a parallel combination of resistors, we have the equivalent resistance of the circuit as
$\Rightarrow \dfrac{\text{1}}{{{\text{R}}_{\text{parallel}}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}+--+\dfrac{1}{{{R}_{n}}}$
Now if the resistances have equal values, we can say that
$\Rightarrow \dfrac{\text{1}}{{{\text{R}}_{\text{parallel}}}}=\dfrac{n}{R}\Rightarrow {{\text{R}}_{\text{parallel}}}=\dfrac{R}{n}$ ,where $n$ is the no. of resistors and $R$ is the value of one resistor
So, for a parallel combination of the electric bulbs, we can say that $\Rightarrow R=\dfrac{4840}{n}\Omega $
Applying ohm’s law, we have
$\Rightarrow V=IR$ where $I$ is the maximum allowable current through the circuit
Substituting the values, we get
$\Rightarrow 220=5\times \dfrac{4840}{n}$
Simplifying this equation, we get
$\therefore n=\dfrac{4840\times 5}{220}=110$
Hence $110$ electric lamps can be connected in the circuit .

Note
Be careful as to not apply the formula $P=VI$ for finding the current through one electric bulb. That method, although correct, should only be applied for resistors connected in series; the current through every element remains the same. For parallel connection, we have a constant voltage through every element so we should proceed accordingly.