Question

A circuit is created with a source and four identical resistors of resistance $R$ and source voltage $V$ as shown above.
The current through $R_3$ is

• A. $\frac{V}{5R}$
• B. $\frac{2V}{5R}$
• C. $\frac{3V}{5R}$
• D. $\frac{4V}{5R}$
• E. $\frac{V}{R}$

Answer 1

The correct option is A $\frac{V}{5R}$

We can see from the given circuit , that $R_3$ and $R_4$ are in series , therefore equivalent resistance

$R^{\prime }=R+R=2R$

now , $R^{\prime }$ and $R_2$ are in parallel , therefore equivalent resistance

$R^{\prime \prime }=\frac{R\times 2R}{R+2R}=2R/3$

since $R_1$ and $R^{\prime \prime }$ are in series , therefore final equivalent resistance

$R^{\prime \prime \prime }=2R/3+R=5R/3$ ,

therefore current in the circuit , $I=V/R^{\prime \prime \prime }=V/\left(5R/3\right)=3V/5R$ ,

now by Kirchhoff's law in closed mesh , having $R_1$ and $R_3,R_4$ ,

$V=I{R}_1+I^{\prime }{R}_3+I^{\prime }{R}_4=3V/5R\times R+2I^{\prime }R$

or $V=3V/5+2I^{\prime }R$

or $2I^{\prime }R=V-3V/5=2V/5$

or $I^{\prime }=V/5R$