Question

When two bulbs of power 60 W and 40 W are connected in series, then the power of their combination will be -

Answer 1

Given : Two bulbs of power 60W and 40W are connected in series.

To find : The power of their combination.

let's derive the formula,

let P₁ and P₂ are the powers of two given bulbs A and B respectively and both are connected in series combination.

resistance of bulb A, R₁ = V²/P₁

resistance of bulb B, R₂ = V²/P₂

we know, equivalent resistance of two resistors which are joined in series combination is given by, Req = R₁ + R₂

⇒V²/Peq = V²/P₁ + V²/P₂

⇒ 1/Peq = 1/P₁ + 1/P₂ , where Peq is power of their series combination.

now using above formula to find power of combination

here P₁ = 60 W, P₂ = 40W

so, 1/Peq = 1/60 + 1/40

= 2/120 + 3/120

= (2 + 3)/120 = 5/120

= 1/24

so, Peq = 24 W

Therefore the power of their combination is 24W

Answer 2

Given :-

Two bulbs of power 60 W and 40 W are connected in series.

To determine :-

The power of their combination.

Acknowledgement :-

We must know,

$\longrightarrow \displaystyle{\sf{P=\frac{V^2}{R} }}$

• Where, P is the power

• V is the potential difference and,

• R is the resistance.

$\longrightarrow$ If resistors are connected in series, then,

$\displaystyle{\sf{R_{Equivalent}=R_1+R_2+R_3+......+R_n}$           ...(i)

Solution :-

As given above,

$\displaystyle{\sf{P=\frac{V^2}{R} }}$, we can also write it as $\displaystyle{\sf{R=\frac{V^2}{P} }}$.

Now, putting this value of R in equation (i) :-

$\displaystyle{\sf{(\frac{V^2}{P} )_{Equivalent}=\frac{V^2}{P_1}+\frac{V^2}{P_2}+\frac{V^2}{P_3}+.....+\frac{V^2}{P_n} }}$

Cancelling V² in both the sides :-

$\displaystyle{\sf{(\frac{1}{P} )_{Equivalent}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+.....+\frac{1}{P_n} }}$

$\rule{120}{2}$

According to the question, there are 2 bulbs with power 60 W and 40 W.

So, from the above equation of Power of combination, we get :-

$\displaystyle{\sf{(\frac{1}{P} )_{Equivalent}=\frac{1}{60}+\frac{1}{40} }}\\\\\\\displaystyle{\sf{\implies (\frac{1}{P} )_{Equivalent}=\frac{2+3}{120} }}\\\\\\\displaystyle{\sf{\implies (\frac{1}{P} )_{Equivalent}=\frac{5}{120} }}\\\\\\\displaystyle{\sf{\implies (\frac{1}{P} )_{Equivalent}=\frac{1}{24} }}$

By reciprocating,

$\large\boxed{\displaystyle{\sf{\implies P_{Equivalent}=24\ W }}}$

Therefore, the power of combination of the two bulbs is 24 Watt.