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?

regard
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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
 
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thanks for complement jegannathan sir
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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
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Hope it is clear
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Why can't we use P=v2/R formula?
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What are you looking for?