Home
Class 12
PHYSICS
A 100 W bulb B1, and two 60 W bulb B2 an...

A 100 W bulb `B_1`, and two 60 W bulb `B_2` and `B_3`, are connected to a 250 V source, as shown in 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_1gtW_2=W_3`

B

`W_1gtW_2gtW_3`

C

`W_1ltW_2=W_3`

D

`W_1ltW_2ltW_3`

Text Solution

Verified by Experts

The correct Answer is:
D
d. We know that `R= V^2/rho`
`:. R_1 = V^2/100, R_2 = V^2/60 = R_3`
`W_1 = (V_(1)^(2))/(R_1) = (V^2R_1)/(R_1+R_2)W_2 = V_(2)^(2)/R_2, W_2 = V_(2)^(2)/R_2`
`W_3:W_2:W_1 : (250)^2/R_3 : (250)^2/(R_1+R_2)^2 R_2 (250)/(R_1+R_2)^2 R_1`
`=60:((250)^2V^2//60)/(V^2/100+V^2/60) : ((250)^2V^2//100)/(V^2/100+V^2/60)`
`=60:25:15`
Hence, `W_3gtW_2gtW_1`.
Promotional Banner

Similar Questions

Explore conceptually related problems

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 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 100 W bulb B_(1) and two 60 W 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

Three bulbs of 40 W , 60 W and 100 W are connected in series with a 240 V source.

When two electric bulbs of 40W and 60W are connected in parallel, then the:

Bulb B_1(100W - 250W) and bulb B_2(100W - 200V) are connected across 250V. What is potential drop across B_2 ?

A 25 W , 220 V bulb and a 100 W , 220 V bulb are connected in parallel across a 440 V line

Two bulbs 100 W, 250 V and 200 W, 250 V are connected in parallel across a 500 V line. Then-

Three electric bulbs of 200 W, 200 W and 400 W are connected as shown in figure. The resultant power of the combination is

Three 60 W , 120 V light bulbs are connected across a 120 V power lines as shown in Fig. 7.26 . Find (a) the voltage across each bulb and (b) the total power dissipated in the three bulbs.