Question

Two electric bulbs p and q have their resistances in the ratio of 1:2. They are connected




in series across a battery. Find the ratio of the power dissipation in these bulbs ?

Answer 1

Answer:

answer is 1:2

Explanation:

As the resistances are connected in series current is same for both .

by formula heat dissipation = i^2 R

where i is the current through resistors.

As current is same for both heat dissipated is directly proportional to resistance.

so the ratio will be 1 :2 as that of resistance.

HOPE it will be helpful

Answer 2

The power dissipation in these bulbs is the equal to the ratio of resistance offered.

1:2

Explanation :

The ratio of the resistance of two electric bulbs P and Q are 1:2.

The current is the same in both the bulbs. Because, Both of bulbs are connected in a series across a battery.

The resistance of bulb P is 1 Ω.

The resistance of bulb Q  is 2 Ω.

Now,

Power will be P

$\implies \sf P' = VI\\ \\ \\ \implies \sf { P' = V \dfrac{Q}{t} = VI }$

We know the Ohm's Law :

$\bigstar\;\boxed{\bf V = IR}}$

So,

⇒ P' = I²R

Where,

• P' = Power.
• I = Current.
• R = Resistance.

Now,

• For bulb P :

⇒ Power = P1

⇒ P1 = I²(1)

• For bulb Q :

⇒ Power = P2

⇒ P2 = I²(2)

Therefore,

$\implies \sf \dfrac{P1}{P2} = \dfrac{(I^2 \times 1)}{(I^2 \times 2)}\\ \\ \\ \implies \sf \dfrac{P1}{P2} = \dfrac{1}{2}$

Hence,

The power dissipation in these bulbs will be equal to the ratio of resistance offered, like 1:2.