Question

two resistors of 4r and 6r are 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

Here's ur answer.. Mark mine as brainliest

Answer 2

★Chapter -Electricity★

★Questions level-Medium★

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

•Two resistors of 4Ω and 6Ω are 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

▬▬▬▬▬▬▬▬▬▬▬▬▬

$\sf\blue\bigstar$ANSWER:-

Here,$\sf R_1$=4Ω and $\sf R_2$=6Ω

Both are in parallel .Hence,

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

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

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

=2.4Ω

i)Power dissipated,

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

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

ii)For $\sf R_1,$ Power dissipated

$\sf \red{P{_1}=\dfrac{V^{2}}{R_{1}}}$

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

=9W

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

For $\sf R_2$ Power dissipated

$\sf \green{P_{2}=\dfrac{V^{2}}{R_{2}}}$

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

=6W

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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

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