Question

Two lamps, one rated 100 W at 220V are connected 1) in series and 2) in parallel to electric main supply of 220 V.

Find the current in each case....

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Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ One lamp is rated 100 W at 220 V and the other is rated 200 W at 220 V

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ The current in each case?

$\displaystyle\large\underline{\sf\gray{Solution}}$

Correct Question

Two lamps, one rated 100 W at 220V and the other rated 200 W at 220 V are connected 1) in series and 2) in parallel to electric main supply of 220 V. Find. the current drawn in each case.

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

So we know that,

$\displaystyle\sf \underline{\boxed{\sf P = \dfrac{V^2}{R}}}$

Which can also be written as,

$\displaystyle\sf \underline{\underline{\sf R = \dfrac{V^2}{P}}}$

When Power is 100 W at 220 V

➝ $\displaystyle\sf R_1 = \dfrac{V^2}{P_1}$

➝ $\displaystyle\sf R_1 = \dfrac{220^2}{100}$

➝ $\displaystyle\sf R_1 = \dfrac{48400}{100}$

➝ $\displaystyle\sf\purple{R_1 = 484 \ \Omega}$

When Power is 200 W at 220 V

»» $\displaystyle\sf R_2 = \dfrac{V^2}{P_2}$

»» $\displaystyle\sf R_2 = \dfrac{220^2}{200}$

»» $\displaystyle\sf R_2 = \dfrac{48400}{200}$

»» $\displaystyle\sf\purple{R_2 = 242 \ \Omega}$

So when they are connected in series then their equivalent resistance will be,

➢ $\displaystyle\underline{\boxed{\sf R_{eq} = R_1+R_2+R_3...R_n}}$

➢ $\displaystyle\sf R_{eq} = 484+242$

➢ $\displaystyle\sf \green{R_{eq} = 726 \ \Omega}$

Similarly when when they are connected in parallel,

➢ $\displaystyle \underline{\boxed{\sf\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}...\dfrac{1}{R_n}}}$

➢ $\displaystyle\sf \dfrac{1}{R_{eq}} = \dfrac{1}{484}+\dfrac{1}{242}$

➢ $\displaystyle\sf \dfrac{1}{R_{eq}} = \dfrac{1}{484}+\dfrac{2}{484}$

➢ $\displaystyle\sf \dfrac{1}{R_{eq}} = \dfrac{3}{484}$

➢ $\displaystyle\sf \green{R_{eq} = \dfrac{484}{3}}$

And so the Current in the circuit will be,

➠ $\displaystyle\sf V = IR$

➠ $\displaystyle\sf \dfrac{V}{R} = I$

Case 1 (Series Connection)

➳ $\displaystyle\sf I_1 = \dfrac{220}{726}$

➳ $\displaystyle\sf \orange{I_1 = 0.30 \ A}$

Case 2 (Parallel Connection)

➳ $\displaystyle\sf I_2 = \dfrac{220\times 3}{484}$

➳ $\displaystyle\sf I_2 = \dfrac{660}{484}$

➳ $\displaystyle\sf \pink{I_2 = 1.36 \ A}$

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Answer 2

Given :-

Resistance of 1st lamp, R₁ = V₁²/P₁ = (220)²/100

= 484 Ω

Resistance of 2nd lamp, R₂ = V₂²/P₂ = (220)²/60

= 806.6 Ω

Solution :-

Lamps are connected in parallel combination,

= R₁R₂/(R₁ + R₂)

= 484 × 806.6/(484 + 806.6)

= 390394.4/1290.6

= 302.5 Ω

Now, Current

I = V/Req

I = 220/(302.5)

I = 0.72 Ampere