Question

Find the e.m.f. (in $V$) ${\epsilon }_1$ and ${\epsilon }_2$ in the circuit as shown in the figure. Also find the potential
difference (in $V$) of point '$a$' relative to point '$b$'.
${\epsilon }_1$ =
${\epsilon }_2$ =
$V_a-V_b$ =

Answer 1

Let $V_b$ and $V_a$ be the potentials at point $b$ and $a$ respectively.
On Applying Kirchhoff's current rule at junction $a$, we can say
$1A+$ current in arm $ab$ $=2A$
$⟹$ current in arm $ab=1A$


Mark points ($c,d,e,f$) on the circuit and apply Kirchhoff's voltage rule in loop ($abcd$)


$⟹1\times 4+1\times 1-{ϵ}_1-6\times 1+20-1\times 1=0$
$⟹4+1-{ϵ}_1-6+20-1=0$
$⟹{ϵ}_1=18V$

Now apply Kirchhoff's voltage law in loop $abfe$
$⟹-2\times 1-{ϵ}_2-2\times 2+{ϵ}_1-1\times 1-1\times 4=0$
$⟹-2-{ϵ}_2-4+18-1-4=0$
$⟹{ϵ}_2=7V$

On further applying Kirchhoff's voltage rule in arm $ab$,we get
$V_b+18-1\times 1-4\times 1=V_a$
$⟹V_a-V_b=13V$