Question

In the arrangement shown in figure, the current through $5\Omega$ resistor is

• A. 2 A
• B. zero
• C. $\frac{12}{7}A$
• D. 1 A

Answer 1

Answer: (A)

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 $E_{eq}=12V$
Hence current through $5\Omega$ resistor
$I=\frac{E_{eq}}{r_{eq}+5}=\frac{12}{1+5}=2A$