A 100v battery is connected to the electric network .If the power consumed in the 2ohm resistor is 200W,determine the power dissipated in the 5ohm resistor?

Hello Pavani, friend Shivam has done an excellent task. Let us congratulate that friend
But we could do the problem even simpler
[First with 6 and 30 ohm being parallel would have 30*6/(30+6) = 5 ohm as effective
This 5 with another 5 in series would have effective 10 ohm (top line)
In the second line we have 40 ohm which becomes parallel to 10 ohm found.
So effective will be 10*40/(10+40) = 8 ohm
This 8 ohm is in series with 2 ohm. Hence effective 10 ohm
So current coming out of battery I = V/R = 100/10 = 10 A]
This 10 A could also be found easily just by taking into account of the power dissipation in 2 ohm.
I = ​√P/R = √200/2 = √100 = 10 A
Now this confirms that 10 A current comes out of the battery
Ratio of resistance in top and second line is 10 : 40 i.e. 1: 4
Hence the current across them will be in the ratio 4 : 1 [recall current and resistance are inversely related]
So 10 A would be divided as 8 A through 10 ohm route and 2 A via 40 ohm route
Thus 5 ohm in the top line has 8 A passing through it
Hence power dissipation in 5 ohm resistor = I^2 * R = 8^2 * 5 = 64 * 5 = 320 W
 

thanks for complement jegannathan sir

First with 6 and 30 ohm being parallel would have 30*6/(30+6) = 5 ohm as effective This 5 with another 5 in series would have effective 10 ohm (top line) In the second line we have 40 ohm which becomes parallel to 10 ohm found. So effective will be 10*40/(10+40) = 8 ohm This 8 ohm is in series with 2 ohm. Hence effective 10 ohm So current coming out of battery I = V/R = 100/10 = 10 A] This 10 A could also be found easily just by taking into account of the power dissipation in 2 ohm. I = ​√P/R = √200/2 = √100 = 10 A Now this confirms that 10 A current comes out of the battery Ratio of resistance in top and second line is 10 : 40 i.e. 1: 4 Hence the current across them will be in the ratio 4 : 1 [recall current and resistance are inversely related] So 10 A would be divided as 8 A through 10 ohm route and 2 A via 40 ohm route Thus 5 ohm in the top line has 8 A passing through it Hence power dissipation in 5 ohm resistor = I^2 * R = 8^2 * 5 = 64 * 5 = 320 W

Hope it is clear

Why can't we use P=v2/R formula?

HELP OTHERS WITH A 😃

Answer 2