Question

two resistors of 20 eaxh are connected in parallel with a battery of 10 total current passing through the circuit is​

Answer 1

$\underline{\underline{\Large{\bf Given:-}}}$

$\bull\sf R_1=R_2=20\Omega$

$\bull\sf Potential\:Difference=V=10V</p><p>[tex]\underline{\underline{\Large{\bf To\:Find:-}}}$

$\bull\sf current=I$

$\underline{\underline{\Large{\bf Given:-}}}$

Firstly lets find Equivalent Resistance

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

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{20}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1+1}{20}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{2}{20}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1}{10}$

$\\ \sf\longmapsto R=10\Omega$

Using ohms law

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

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

$\\ \sf\longmapsto I=\dfrac{10}{10}$

$\\ \sf\longmapsto I=1A$

$\underline{\underline{\therefore{\sf Current\:flowing\:through\:the\:circuit\;is\:1A.}}}$