Question

two electric bulb P and Q have their resistance are in 1:2.they are connected in series across a battery. Find the power dissipation in these bulb
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Answer 1

Given that, the ratio of the resistance of two electric bulbs P and Q are 1:2.

Both of them (bulbs) are connected in a series across a battery. Means current is the same in both the bulbs.

The resistance of bulb P, resistance is 1Ω and for bulb Q, resistance is 2Ω.

Now,

P' = VI

{ P' = V (Q/t) = VI }

Here, Power is denoted by P'.

Also, from Ohm's Law

V = IR

So, P' = I²R

{ P' = Power, I = Current and R = Resistance }

For bulb P: (Power = P1)

→ P1 = I²(1)

For bulb Q: (Power = P2)

→ P2 = I²(2)

So,

→ P1/P2 = (I² × 1)/(I² × 2)

→ P1/P2 = 1/2

The power dissipation in these bulbs is the same or equal to the ratio of resistance offered i.e. 1:2.

Answer 2

Given:-

$\rightarrow \texttt{Two electric bulbs P and Q have their resistance in the ratio}\\\texttt{ of 1 : 2.}$

$\rightarrow\texttt{They are connected in series across a battery.}$

To find:-

$\texttt{The power dissipation in the two bulbs.}$

Solution:-

$\texttt{Resistance of bulb P is }$ 1 Ω, $\texttt{and the resistance of bulb Q is }$ 2Ω$\texttt{.}$

$\texttt{Current across both the bulbs is same, as they are connected}\\\texttt{in series.}$

$\texttt{Now, we know,}\\\\\displaystyle\sf\bold{{{Power(P)=I^2R}}}$

$\texttt{So, we can write that Power of P/Power of Q = } \sf{I^2R_1/I^2R_2.}\\\\R_1\texttt{is the resistance of bulb P, and }R_2\ \texttt{is the resistance of bulb Q.}$

$\displaystyle{\sf{\frac{P_1}{P_2}=\frac{I^2R_1}{I^2R_2} }}\\\\\displaystyle{\sf{\implies \frac{P_1}{P_2}=\frac{R_1}{R_2} }}\\\\\underline{\texttt{P}_1\texttt{ is the power of bulb P and P}_2\texttt{ is the power of bulb Q.}}\\\\\underline{\texttt{I}^2\texttt{ gets cancelled in both, denominator and numerator because}}\\\underline{\texttt{current is same through both the bulbs.}}$

$\boxed{\displaystyle{\sf{\implies \frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{1}{2} } }}$

$\texttt{As R}_1\texttt{:R}_2\texttt{ = 1:2, then }\frac{R_1}{R_2}=\frac{1}{2}.$