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Q.
In the arrangement shown in figure, the current through $5\, \Omega$ resistor is
Current Electricity
Solution:
In this circuit there are 2 cells connected in parallel Equivalent internal resistance,
$r _{ eq }=\frac{ r _{1}+ r _{2}}{ r _{1} r _{2}}$
where $r_{1}$ and $r_{2}$ are the respective internal resistance hence
$r _{ eq }=1\, \Omega$
And Equivalent e.m.f, $\frac{ E _{ eq }}{ r _{ eq }}=\frac{ E _{1}}{ r _{1}}+\frac{ E _{2}}{ r _{2}}$
Hence on solving we get $E _{ eq }= 1 2\, V$
Hence current through $5 \Omega$ resistor
$I =\frac{ E _{ eq }}{ r _{ eq }+5}=\frac{12}{1 +5}= 2\, A$
