two resistors of 4 ohm and 6 ohm are connected in parallel.the combination is connected across a 6 V battery of negligence resistance calculate(a)the power supplied by the battery (b)the power dissipated in each resistor

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Solution

$E q u i v a l e n t R e s i s \tan c e, R' = \frac{4 \times 6}{4 + 6} = 2.4 o h m I = \frac{V}{R} = \frac{6}{2.4} = 2.5 P = I^{2} R = 2 . 5^{2} \times 2.4 = 15 P o w e r d i s s i p a t i o n, P = \frac{V^{2}}{R} = \frac{6^{2}}{4} = 9 W P = \frac{6^{2}}{6} = 6 W$

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two resistors of 4 ohm and 6 ohm are connected in parallel.the combination is connected across a 6 V battery of negligence resistance calculate(a)the power supplied by the battery (b)the power dissipated in each resistor

Equivalent Resistance,R'=4×64+6=2.4ohmI=VR=62.4=2.5P=I2R=2.52×2.4=15Power dissipation,P=V2R=624=9WP=626=6W

$E q u i v a l e n t R e s i s \tan c e, R' = \frac{4 \times 6}{4 + 6} = 2.4 o h m I = \frac{V}{R} = \frac{6}{2.4} = 2.5 P = I^{2} R = 2 . 5^{2} \times 2.4 = 15 P o w e r d i s s i p a t i o n, P = \frac{V^{2}}{R} = \frac{6^{2}}{4} = 9 W P = \frac{6^{2}}{6} = 6 W$