Physics, asked by TrustedAnswerer19, 5 hours ago


 \sf \: in \: the \:  \: figure \:  \\  \\  \sf \: voltage \:  \: V_1 = 35\: V \\  \bf \: registance \: R_1 = 10 \: \Omega \:  \\  \bf \: registance \: R_2 = 15 \: \Omega \:  \\  \bf \: registance \: R_ 3 = 6 \: \Omega \:  \\  \bf \: registance \: R _ 4= 20 \: \Omega \:  \\  \bf \: registance \: R_5 = 25 \: \Omega \:  \\  \bf \: registance \: R_6= 5 \: \Omega \:  \\  \bf \: registance \: R_7= 12\: \Omega \:  \\  \bf \: registance \: R_8 = 15 \: \Omega \:  \\    \\   \green \odot \: \sf \: find  \: the \: value \: of \: \: total \: current \:  (I )\: \\

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Answered by NewGeneEinstein
1

Answer:

Here

R1 ,R2,R7,R8 are in series.

R3,R4,R5,R6 are in parallel.

Lets calculate effective or equivalent resistance

in Parallel:-

\boxed{\sf \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\dots}

\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{20}+\dfrac{1}{25}

\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{60+50+15+12}{300}

\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{137}{300}

\\ \sf\longmapsto R=\dfrac{300}{137}

\\ \sf\longmapsto R=2.1\Omega

In series:-

\boxed{\sf R=R_1+R_2+\dots}

\\ \sf\longmapsto R=10+15+15+12

\\ \sf\longmapsto R=52\Omega

Now Equivalent resistance:-

\\ \sf\longmapsto R_{eq}=52+2.1

\\ \sf\longmapsto R_{eq}=54.1\Omega

  • Potential difference=V=35V

Using ohms law

\boxed{\sf \dfrac{V}{I}=R}

\\ \sf\longmapsto I=\dfrac{V}{R}

\\ \sf\longmapsto I=\dfrac{35}{54}

\\ \sf\longmapsto I=0.6A

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