Question

Three balls are 25 Watt ,40W , 100W respectively are connected in series and parallel to a voltage source which physical quantity is same for the bulbs in two arrangements

Answer 1

Answer:

thus , R= V^2/P

Let bulbs of 25W, 40W , and 60W have resistance R(1) ,R (2) and R (3)

R (1)= 6*6/25 =1.44 ohm

R (2) = 6*6/40=0.9 ohm

R (3)= 6*6/60=0.6 ohm

The bulb with highest resistance will glow the brightest, thus bulb with 25 W is the brightest

Answer is 25 W

Answer 2

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Three balls are 25 Watt ,40W , 100W respectively are connected in series and parallel to a voltage source. Which physical quantity is same for the bulbs in two arrangements ?

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➡ At R2 & R3

$\mathtt{\huge{\underline{\orange{Solution:-}}}}$

Given :-

• Three balls are 25 Watt ,40W , 100W respectively are connected in series and parallel to a voltage source.

To Find :-

• When the physical quantity is same for the bulbs in two arrangements.

Calculation :-

According to the question,

Resistance is the property by virtue of which the current opposes to flow through a conductor.

R = V/I ....... (1)

I = P/V.........(2)

From 1 & 2

R = V/I/P/V

R= V²/P

Let bulbs of 25W, 40W , and 60W have , resistance R(1) ,R (2) and R (3)

• R (1)= 6×6/25 = 36/25 = 1.44 Ω

• R (2) = 6×6/40= 36/40 = 0.9 Ω

• R (3)= 6×6/60= 36/60 = 0.7 Ω

At R2 & R3 ; R2≈R3

The physical quantity is same for the bulbs in two arrangements at R2&R3.

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