Question

If the wire has resistivity $\rho$ and cross sectional area A, the equivalent resistance between P and Q is?

• A. $\frac{\rho l}{\sqrt{2}A}$
• B. $\frac{\sqrt{2}\rho l}{A}$
• C. $\frac{2\rho l}{A}$
• D. $\frac{\rho l}{A}$

Answer 1

The correct option is A $\frac{\rho l}{\sqrt{2}A}$

We need to find the total length of the wire through which we can calculate the resultant resistance.

In the figure, $V_A=V_F$ {Some potential }

$\therefore ⟹$ that there be no current through arm $AF$ and hence we can calculate that, similarly there will be no current through arm $CD$ because of the same potential drop. Hence, we can also calculate that part.

Now, we will calculate the resistance of each arm,

$R_{PA}=\frac{\rho l}{2A}=R_{PF}$

Also, resistance $ABC$ is in parallel connection with resistance $AC$.

$\frac{1}{R_1}=\frac{1}{R_{ABC}}==\frac{1}{R_{AC}}=\frac{A}{\rho l}+\frac{\sqrt{2}A}{\rho l}$

$=\frac{A(\sqrt{2}+1)}{\rho l}$

$⟹R_1=\frac{\rho l}{A(\sqrt{2}+1)}$

As we can deduce from the figure, that resistance due to $PA,(ABC,AC),C\theta$ are in series hence, their total ressitance is given by,

$R_2=R_{PA}+R_1+R_{C\theta }=\frac{\rho l}{2A}+\frac{\rho l}{A(\sqrt{2}+1)}+\frac{\rho l}{2A}$

$=\frac{\rho l}{A}(1+\frac{1}{\sqrt{2}+1})$

$=\frac{\rho l}{A}\left(\frac{\sqrt{2}+2}{\sqrt{2}+1}\right)$

Similarly, we will calculate the resistance for part $PFED\theta$ which will be same as $R_2$

Now, $R_2$ and $R_3$ are resistance in parallel

$⟹\frac{1}{R_{eq}}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{2}{R_2}$

$R_{eq}=\frac{R_2}{2}=\frac{\rho l}{2A}\left(\frac{\sqrt{2}+2}{\sqrt{2}+1}\right)$

$R_{eq}=\frac{\rho l}{\sqrt{2}A}\Omega$