A battery supplies 150W and 196W power to two resistors of 6 ohms and 4 ohms when they are connected separately to it. The initial resistance of the battery is

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Solution

$W h e n b a t t e r y o f r e s i s \tan c e R i s c o n n e c t e d t o 6 o h m' s r e s i s \tan c e t h e n; P_{1} = V \times I = V^{2} / (R + 6) a n d P_{2} = V \times I = V^{2} / (R + 4) T h e r e f o r e \frac{P_{1}}{P_{2}} = \frac{(R + 4)}{(R + 6)} o r \frac{150}{196} = \frac{(R + 4)}{(R + 6)} o r 150 R + 900 = 196 R + 784 o r 46 R = 116 o r R = 116 / 46 = 2.52 o h m T h e r e f o r e t h e i n i t i a l r e s i s \tan c e o f t h e b a t t e r y i s 2.52 o h m .$

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A battery supplies 150W and 196W power to two resistors of 6 ohms and 4 ohms when they are connected separately to it. The initial resistance of the battery is

When battery of resistance R is connected to 6ohm's resistance then; P1=V×I=V2/R+6and P2=V×I=V2/R+4Therefore P1 P2=R+4R+6or 150196=R+4R+6or 150R+900=196R+784or 46R=116or R=116/46=2.52ohmTherefore the initial resistance of the battery is 2.52ohm.