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Question: Two lamps one rated \[60W,\;220V\] and the other \[40W,\;220V\]are connected in parallel to a \[220V...

Two lamps one rated 60W,  220V60W,\;220V and the other 40W,  220V40W,\;220Vare connected in parallel to a 220V220V electric supply mains. What is the total current drawn from the electric mains if the voltage of electric supply is220V220V?

Explanation

Solution

We should check whether the bulbs are arranged in series or in parallel. We should know that voltage will be constant in a parallel circuit and current is constant in series circuit.


Complete step by step answer:
Parallel circuit is a closed circuit in which the current divides into two or more paths before recombining to complete the circuit.In parallel, both bulbs have the same voltage across them. The bulb with the lower resistance will conduct more current and therefore have a higher power dissipation and brightness. Most household electrical wiring bulbs are wired in parallel.

We have given two lamps in such way:
Power of Ist{I^{st}} lamp, P1=40W{P_1} = 40W
Voltage of Ist{I^{st}}lamp, V1=220V{V_1} = 220V.
Power of 2nd{2^{nd}}lamp, P2=60W{P_2} = 60W,
Voltage of 2nd{2^{nd}}lamp V2=220V{V_2} = 220V
We know one thing.
Power = V2R{\text{Power = }}\dfrac{{{V^2}}}{R} [When potential difference is same then considerP=V2RP = \dfrac{{{V^2}}}{R}]
So, resistance of Ist{I^{st}}lamp, R1=V12P1{R_1} = \dfrac{{V_1^2}}{{{P_1}}}
=(220)260=4840060 =806.67Ω  = \dfrac{{{{\left( {220} \right)}^2}}}{{60}} = \dfrac{{48400}}{{60}} \\\ = 806.67\Omega \\\
Resistance of 2nd{2^{nd}}lamp, R2=V22P2{R_2} = \dfrac{{V_2^2}}{{{P_2}}}

=(220)240=4840040 =1210Ω  = \dfrac{{{{\left( {220} \right)}^2}}}{{40}} = \dfrac{{48400}}{{40}} \\\ = 1210\Omega \\\

Now question said,
Both lamps are parallel.
So,
1Req=1R1+1R2 1Req=1806.67+11210 1Req=1210+806.671210×806.67  \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \\\ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{806.67}} + \dfrac{1}{{1210}} \\\ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{{1210 + 806.67}}{{1210 \times 806.67}} \\\
Reff=484Ω{R_{eff}} = 484\Omega
Now, current drawn from electrical supply, i = potential difference/Req.
=220V484=2044=511= \dfrac{{220V}}{{484}} = \dfrac{{20}}{{44}} = \dfrac{5}{{11}}

So, the correct answer is Itotal=511Amp = 0.4545Amp{I_{total}} = \dfrac{5}{{11}}{\text{Amp = 0}}{\text{.4545Amp}}.

Note:
In a parallel circuit, charge is divided up into separate branches such that there can be more current in one branch than there is in another. Nonetheless when taken as a whole, the total amount of current in all the branches when added together is the same as the amount of current at locations outside the branches.