Question

Two resistors A and B of resistance 4 ohm and 6 ohm respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate : the power supplied by the battery ,and the power dissipated in each resistor.

Answer 1

P=V^2/R
P=6^2/10
=36/10=3.6watt(power by battery)
In first resistance 4/6=0.6A
In second resistance 1A.
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Answer 2

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

Two resistors A and B of resistance 4 ohm and 6 ohm respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate :

i)the power supplied by the battery ,and

ii)the power dissipated in each resistor.

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$\sf\red\bigstar$ANSWER:-

Here,$\sf R_1$=4Ω and $\sf R_2$=6Ω

Both are in parallel .Hence,

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

$\sf{\red\bigstar \blue{R_{eq}}=\orange{\dfrac{R_1 R_2}{R_1+R_2}}}$

$\sf =\dfrac{4\times 6}{4+6}$

=2.4Ω

i)Power dissipated,

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

$=\sf\dfrac{36\times 10}{24}=15W$

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ii)For $\sf R_1,$ Power dissipated

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

$\sf =\dfrac{\cancel{36}\:9}{\cancel{4}}$

=9W

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For $\sf R_2$ Power dissipated

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

$\sf =\dfrac{6\times \cancel{6}}{\cancel{6}}$

=6W

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MORE TO KNOW

✦Resistor in series =$\bf {\red{\boxed{\sf R_S=R_1+R_2+R_3.....}}}$

✦Resistor in parallel =$\bf \purple{\boxed{\sf \dfrac{1}{R_{p}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}........}}$