Question

In the circuit shown PD across R is V. Find the PD across resistance 2R.

Answer 1

Let the supply through the whole circuit be E

Then current through the whole circuit will be:

$I=\frac{E}{R_{eq}}$($R_{eq}=$ equivalent resistance of the circuit)

As $2R$ and $4R$ are in series their equivalent resistance $=2R+4R=6R$

Then $6R$ and $3R$ are in parallel so their equivalent $=\frac{6\times 3}{6+3}=\frac{18}{9}R=2R$

Again R and $2R$ are in series, so $R_{eq}=R+2R=3R$

So $I=\frac{E}{3R}$

Hence potential drop across R will be $V=IR=\frac{E}{3R}\times R=\frac{E}{3}$

$\Rightarrow E=3V$

Current through $2R$ and $4R$ branch $i=\left(\frac{3V-V}{2R}\right)\left(\frac{3R}{3R+6R}\right)$ (current division rule)

$\Rightarrow i=\frac{V}{3R}$

So potential drop across resistance $2R$ is: $V_{2R}i\left(2R\right)=\frac{V}{3R}\left(2R\right)=\frac{2V}{3}$.