Question

Two resistors of 3 ohm and 9 ohm are connected in parallel. The combination is connected across a 9 V battery of negligible resistance. Calculate the power supplied by the battery and the power dissipated in each resistor

Answer 1

1 / R = 1 / 3 + 1 / 9

1 / R = 4 / 9

R = 9/ 4

V = IR

9 = I ( 9 /4 )

I = 4 ampere

P =VI

P = 9× 4 = 36 W

Answer 2

$\sf\red{\bigstar}$QUESTION:-

•Two resistors of 3 ohm and 9 ohm are connected in parallel. The combination is connected across a 9 V battery of negligible resistance. Calculate the power supplied by the battery and the power dissipated in each resistor

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$\sf\red\bigstar$ANSWER:-

Here,$\sf R_1$=3Ω and $\sf R_2$=9Ω

Both are in parallel .Hence,

$\sf\pink\bigstar{ \pink{\dfrac{1}{R_{eq}}}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}=\dfrac{R_2+R_{1}}{R_{1} R_{2}}$

$\sf{\pink\bigstar R_{eq}}=\dfrac{R_1 R_2}{R_1+R_2}$

$\sf =\dfrac{3\times 9}{3+9}$

=2.25Ω

i)Power dissipated,

$\sf P=\dfrac{V^{2}}{R}$

$=\sf\dfrac{81\times 100}{225}=36W$

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ii)For $\sf R_1,$ Power dissipated

$\sf P_1=\dfrac{V^{2}}{R_{1}}$

$\sf =\dfrac{\cancel{81}\:27}{\cancel{3}}$

=27W

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For $\sf R_2$ Power dissipated

$\sf P_2=\dfrac{V^{2}}{R_{2}}$

$\sf =\dfrac{9\times \cancel{9}}{\cancel{9}}$

=9W

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MORE TO KNOW

•P=VI=I²R=V²/R

•Resistor in series =$\bf R_S=R_1+R_2+R_3.....$

•Resistor in parallel =$\bf \dfrac{1}{R_{p}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}........$