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 power dissipation in these bulbs​

Answer 1

Given :

Two electric bulbs P and Q have their resistances in the ratio of 1:2. Both are connected in series across a battery.

To Find :

Ratio of power dissipation in these bulbs.

Solution :

❖ When the bulbs (resistances) are connected in series, the current I through each resistance is same. Consequently

• P ∝ R (∵ P = I²R)

Hence in a series combination of resistances, the potential difference, power consumed and hence heat produced will be larger in the higher resistance.

$\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{R_1}{R_2}$

$\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{1}{2}$

$:\implies\:\underline{\boxed{\bf{\purple{P_1:P_2=1:2}}}}$

Knowledge BoosteR :

• When the resistances are connected in parallel, the potential difference V is same across each resistance. Consequently, P ∝ 1/R
• Hence in a parallel combination of resistances, the current, power consumed and hence heat produced will be larger in the smaller resistance.

Answer 2

Given :

Two electric bulbs P and Q have their resistances in the ratio of 1:2. Both are connected in series across a battery.

To Find :

Ratio of power dissipation in these bulbs.

Solution :

❖ When the bulbs (resistances) are connected in series, the current I through each resistance is same. Consequently

P ∝ R (∵ P = I²R)

Hence in a series combination of resistances, the potential difference, power consumed and hence heat produced will be larger in the higher resistance.

$\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{R_1}{R_2}$

$\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{1}{2}$

$:\implies\:\underline{\boxed{\bf{\purple{P_1:P_2=1:2}}}}$

Knowledge BoosteR :

When the resistances are connected in parallel, the potential difference V is same across each resistance. Consequently, P ∝ 1/R

Hence in a parallel combination of resistances, the current, power consumed and hence heat produced will be larger in the smaller resistance.