Q. The potential difference between the ends of $4\, \Omega$ resistance in the given circuit is

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A

$1.2\, V$

B

$2.6\, V$

C

$6.4\, V$

D

$4.8 \,V$

Solution:

we have to find the main current =current through $4$ ohm $=$ I
We know that branch current
$l_{1}=\mid x R_{2} / R_{1}+R_{2}$
$0.8=1 \times 6 / 6+3$
or $0.8=21 / 3=1.2 A$
hence p.d. across $4 \,\Omega= IR$
$=1.2 \times 4=4.8\, V$

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