Two resistors of 60 W and 90 W , respectively, when each connected separately to 120 V source of meg EMF , find the power required in each when they are connected in series with same emf.
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Heya......!!!!
We first need to find the resistance of each resistor .
From P = V²/R
we have R1 = 120²/ 60 = 240 Ω
And R2 = 120²/ 90 = 160 Ω
So ,, when the resistor are connected in series, the current through EMF is same and equal to I = EMF / R1+R2 =>>>. 120/400
=>> 0.3 A .
The Power required to each resistor is
P1 => i²R1 =>> 21.6 W
P2 =. I² R2 =>> 14.4 W .
Hope it helps u. ^ _ ^
We first need to find the resistance of each resistor .
From P = V²/R
we have R1 = 120²/ 60 = 240 Ω
And R2 = 120²/ 90 = 160 Ω
So ,, when the resistor are connected in series, the current through EMF is same and equal to I = EMF / R1+R2 =>>>. 120/400
=>> 0.3 A .
The Power required to each resistor is
P1 => i²R1 =>> 21.6 W
P2 =. I² R2 =>> 14.4 W .
Hope it helps u. ^ _ ^
Donboy1:
Thanks
lol .... blocked....so what? haha, stupid go to hell
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