Question 110 of 380

The current flowing through ${R}_{2}$ is :

A$\dfrac{2}{3}A$B$\dfrac{1}{2}A$C$\dfrac{1}{3}A$D$\dfrac{1}{4}A$

Show answer and solution▾

Answer: (C) $\dfrac{1}{3}A$

As can be seen from the first circuit above, CB&AB are in series which is in parallel with $CA$. So, the net resistance is
$R'=\dfrac{6}{2}\Omega =3\Omega$ as shown in the second diagram.
From the second diagram, DC&CA is in series which is parallel to $DA$. Hence the resistance is
$R''=\left(\dfrac{6\times 3}{(6+3)}\right)\Omega =2\Omega$ as shown in third diagram.
From third diagram, both ED&DA are in series which is parallel to $EA$.
The equivalent resistance as can be seen from the above simplified circuit is ${R}_{eq}=2\Omega$.
The current $I$ through battery, $I=$ $\dfrac{8}{2}=4A$.
From the second diagram, the part $EDACD$ has a resistance of $4\Omega$, the total current will divide equally between EDACD&EA. So consider ${I}_{1}=2A$ flowing through $ED$. Note that ${I}_{1}$ will be the same current flowing in the first diagram through $ED$.
From first diagram, when the resistances are in parallel the current is divided in the inverse ratio. Thus,
$I'$ through $CD=\dfrac{3}{9}\times 2=\dfrac{2}{3}A$
$I''$ through ${R}_{2}=\dfrac{1}{2}\left(\dfrac{2}{3}\right)=\dfrac{1}{3}A$

JEE Main Physics practice question on Current Electricity (Combination of Resistances), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.