Question

Two lamps rated 100v at 220w and 200v at 220w are connected (i) in seriesand (ii) in parallel to electric mains supply of220v. find the current drawn in each case
.

Answer 1

Answer:

The ratings of the lamp are as follows

• 100 V at 220 W
• 200 V at 220 W

They are connected in

• Series
• Parallel

To find:

Current drawn in each case .

Concept:

First you need to find out the Resistance of both the bulbs and place the resistance in simple circuits.

After that , the current drawn can be easily calculated.

Calculation:

For Lamp 1 :

$R1 = \dfrac{ {V}^{2} }{P}$

$= > R1 = \dfrac{ {(100)}^{2} }{220}$

$= > R1 = \dfrac{ 10000 }{220} \Omega$

For Lamp 2

$R2 = \dfrac{ {V}^{2} }{P}$

$= > R1 = \dfrac{ {(200)}^{2} }{220}$

$= > R1 = \dfrac{ 40000 }{220} \Omega$

Now the net resistance in series conditions is :

$= > R_{eq} = ( \dfrac{ 40000 }{220}+ \dfrac{10000}{220} ) \Omega$

$= > R_{eq} = \dfrac{ 50000 }{220} \Omega$

So current drawn in series combination :

$I = \dfrac{200}{ (\frac{50000}{220}) } = 0.88 \: amp$

Net resistance in Parallel combination:

$= > R_{eq} = \dfrac{( \dfrac{40000}{220} \times \dfrac{10000}{220}) }{( \dfrac{40000}{220} + \dfrac{10000}{220}) } \Omega$

$= > R_{eq} = 36.32\Omega$

So current drawn in Parallel combination

$I = \dfrac{200}{ 36.32 } = 5.506 \: amp$

Answer 2

QUESTION:-

Two lamps rated 100v at 220w and 200v at 220w are connected (i) in seriesand (ii) in parallel to electric mains supply of220v. find the current drawn in each case.

Solution :-

IN A FIRST LAMP.

➥R1 = v²/p

➥ R1 = (100)² / 220

➥ R1 = 10000 /220Ω ------(1)

IN A SECOND LAMP.

➥ R2= v² / p

➥ R2 = (200)²/220

➥ R2 = 40000/220Ω----(2)

NOW,

➥ the net resistance in series.

➥ .°. R(eq) = r1+r2

➥.°. R(eq) = (10000 /220 + 40000/220)Ω ---(from 1 & 2)

➥ .°. R(eq) = 50000/220Ω

NOW WE, CALCULATE CURRENT DRAWN IN SERIES COMBINATION.

➥I = V/R(eq)

➥ I = 200/50000/220

➥ .°. I = 0.88A.

NOW WE, CALCULATE NET RESISTANCE IN PARALLEL COMBINATION.

➥ R(eq) = (40000/220 ×10000/220)/(40000/220 +10000/220)Ω

➥ R(eq) = 36.32Ω (ANSWER).

➥ .°.current drawn in parallel combination is ,

➥ I = V/R(eq)

➥I = 220/36.32

➥ .°. I =5.506A (ANSWER) .