A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V sourc

Discussion Forum : Current Electricity

Question -

A 100 W bulb $B_{1}$ and two 60 W bulbs $B_{2}$ and $B_{3}$ are connected to a 250 V source as shown in figure. $W_{1}$ , $W_{2}$ and $W_{3}$ are the output powers of the bulbs $B_{1}$ , $B_{2}$ and $B_{3}$ respectively then

Options:

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: Option D
:
D

The resistances of bulbs $B_{1}, B_{2}$ and $B_{3}$ respectively are
$R_{1} \frac{v^{2}}{w_{1}} = \frac{(250)^{2}}{100} = 625$ ohm
$R_{2} = \frac{(250)^{2}}{60} = 1042$ ohm = $R_{3}$
Voltage across $B_{3}$ is $V_{3} = 250 V$
Voltage across $B_{1}$ is $V_{1} = \frac{V R_{1}}{(R_{1} + R_{2})} = \frac{250 \times 625}{(625 + 1042)} = 93.7 V$
Voltage across $B_{2}$ is $V_{2} = 250 - 93.7 = 156.3 V$
Power output $W_{1} = \frac{V^{2}}{R_{1}} = \frac{(93.7)^{2}}{625} = 14 W$
$W_{2} = \frac{(156.3)^{2}}{1042} = 23 w$
$W_{3} = \frac{(250)^{2}}{1042} = 60 w$
Hence $W_{1} < W_{2} < W_{3}$

Was this answer helpful ?

Next Question

Discussion Forum : Current Electricity

Submit Your Solution hear: