Question

$\sf \: in \: the \: \: figure \: \\ \\ \sf \: voltage \: \: V_1 = 35\: V \\ \\ \bf \: resistance \: R_1 = 10 \: \Omega \: \\ \bf \: resistance \: R_2 = 15 \: \Omega \: \\ \bf \: resistance \: R_ 3 = 6 \: \Omega \: \\ \bf \: resistance \: R _ 4= 20 \: \Omega \: \\ \bf \: resistance \: R_5 = 25 \: \Omega \: \\ \bf \: resistance \: R_6= 5 \: \Omega \: \\ \bf \: resistance \: R_7= 12\: \Omega \: \\ \bf \: resistance \: R_8 = 15 \: \Omega \: \\ \\ \green \odot \: \sf \: find \: the \: value \: of \: \: total \: current \: (I)$
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Answer 1

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

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

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

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

$\\ \sf\longmapsto I=0.6A$