Question

Two metallic wires A and B of the same material are connected in parallel. Wire A has length $l$ and radius $r$,
wire B has a length $2l$ and radius $2r$.Compute the ratio of the total resistance of parallel combination and the
resistance of wire A.

Answer 1

Step 1: Resistance,

Resistance is defined as the property of the conductor that opposes the flow of the current through the conductor. The unit of the resistance is $\Omega$.

The length of wire A is $l$ and radius is $r$.

The length of wire B is $2l$ and radius is $2r$.

The expression to calculate the resistance of the wire is given by

$R=\frac{\rho l}{A}$

Here is $\rho$is the resistivity,$l$ is the length and $A$ is the cross-sectional area of the wire.

The cross-sectional area of the wire is given by

$A=\pi R^2$

Step 2: The resistance of wires,

The resistance for wire A is given by

$R_A=\frac{\rho l}{{\mathrm{\pi r}}^2}$

The resistance for the wire B is given by

$$\begin{gathered} R_B=\frac{\rho \times 2l}{\mathrm{\pi }{\left(2\mathrm{r}\right)}^2} \\ {R}_B=\frac{\rho \times 2l}{4\mathrm{\pi }{\mathrm{r}}^2} \\ {R}_B=\frac{\rho \times l}{2{\mathrm{\pi r}}^2} \end{gathered}$$

Step 3: Calculate the equivalent resistance,

Both the wire are connected in parallel, The resistance is given by

$\frac{1}{R}=\frac{1}{R_A}+\frac{1}{R_B}$

Substitute the expression and solve as

$$\begin{gathered} \frac{1}{R}=\frac{1}{{\displaystyle \frac{\rho l}{{\mathrm{\pi r}}^2}}}+\frac{1}{{\displaystyle \frac{\rho l}{2{\mathrm{\pi r}}^2}}} \\ \frac{1}{R}=\frac{{\mathrm{\pi r}}^2}{\rho l}+\frac{2{\mathrm{\pi r}}^2}{\rho l} \\ \frac{1}{R}=\frac{3{\mathrm{\pi r}}^2}{\rho l} \\ R=\frac{1}{3}\times \frac{\rho l}{{\mathrm{\pi r}}^2} \end{gathered}$$

Substitute the $R_A$ for $\frac{\rho l}{{\mathrm{\pi r}}^2}$in above equation,

$$\begin{gathered} R=\frac{1}{3}\times R_A \\ \frac{R}{R_A}=\frac{1}{3} \end{gathered}$$

Hence, the ratio between total resistance and resistance of the wire A is $1:3$.