Question

A $100W$ bulb $B_1$ and two $60W$ bulbs $B_2$ and $B_3$ are connected to a $250V$ source as shown in the figure. Now $W_1,W_2$ and $W_3$ are the output powers of the bulbs $B_1,B_2$ and $B_3$ respectively. Then

• A. $W_1>W_2=W_3$
• B. $W_1>W_2>W_3$
• C. $W_1<W_2=W_3$
• D. $W_1<W_2<W_3$

Answer 1

Answer: (D)

Let suppose the resistance of bulb $B_1,B_2$ and $B_3$ are $R_1,R_2$ and $R_3$ respectively.
Now by using the formula of power $P=\frac{V^2}{R}$ , We can write $R_1=\frac{V^2}{100},$ $R_2=\frac{V^2}{60}$ and $R_3=\frac{V^2}{60}$ .
Let us suppose that current through series combination of $B_1$ and $B_2$ is $I$ and current through $B_3$ is $I^{{}^{\prime }}$ .
Now by using current division law we can write $I=\frac{E{R}_3}{R_1+R_2+R_3}$ and $I^{{}^{\prime }}=\frac{E\left(R_1+R_2\right)}{R_1+R_2+R_3}$ where $E=250V$
Output power, $W=I^{2}R$
Hence $W_1˸W_2˸W_3=I^2{R}_1˸I^2{R}_2˸{\left(I^{{}^{\prime }}\right)}^2{R}_3$ $\Rightarrow W_1˸W_2˸W_3=\frac{E^2{\left(R_3\right)}^2}{{\left(R_1+R_2+R_3\right)}^2}{R}_1˸\frac{E^2{\left(R_3\right)}^2}{{\left(R_1+R_2+R_3\right)}^2}{R}_2˸\frac{E^2{\left(R_1+R_2\right)}^2}{{\left(R_1+R_2+R_3\right)}^2}{R}_3$
$\Rightarrow W_1˸W_2˸W_3=R_3{R}_1˸R_3{R}_2˸{\left(R_1+R_2\right)}^2$
$\Rightarrow W_1˸W_2˸W_3=\left(\frac{V^2}{60}\times \frac{V^2}{100}\right)˸\left(\frac{V^2}{60}\times \frac{V^2}{60}\right)˸{\left(\frac{V^2}{100}+\frac{V^2}{60}\right)}^2$
$\Rightarrow W_1:W_2:W_3=0.0001:0.0002:0.0007$
$\Rightarrow W_1<W_2<W_3$