Question

The current $I_1$ (in $A$) flowing through $1\Omega$ resistor in the following circuit is

• A. $0.5$
• B. $0.25$
• C. $0.2$
• D. $0.4$

Answer 1

The correct option is C $0.2$

Let the current flowing through the upper branch be $i$.
Given circuit can be reduced as follows:

Since two $1\Omega$ resistances were connected in parallel, their equivalent resistance

$=\frac{1}{2}=0.5\Omega$

So the equivalent resistance in upper branch becomes $R_{eq}=2\Omega +0.5\Omega =2.5\Omega$

The potential difference across upper branch $V=1V$ as it is directly connected across the cell

(since potential difference remains same across parallel branches )

Thus, current flowing through the upper branch is :
$\Rightarrow i=\frac{V}{R_{eq}}$

$=\frac{1V}{2.5\Omega }=0.4A$
(ohm’s law )

Potential difference across parallel combination $0.5\Omega$
$=iR=0.5\times 0.4=0.2V$
(ohm’s law )

this is potential difference across each $1\Omega$
resistor in parallel
current through $1\Omega$ resistor

$I_1=\frac{\left(0.2V\right)}{1\Omega }=0.2A$
(ohm's law )

Hence, option (b) is correct.