Question 119: Current Electricity

Question

A wire of uniform resistance $\lambda \Omega / \mathrm{m}$ is bent into a circle of radius $r$ and another piece of wire with length $2 r$ is connected between points $A$ and $B(\mathrm{AOB})$ as shown in figure. The equivalent resistance between points $A$ and $B$ is $\_\_\_\_$ $\Omega$.

Accepted Answer

For the equivalent resistance between points A and B we can identify three distinct paths for the current to flow from node A to node B. These paths are in parallel.
The resistance per unit for the wire is $\lambda \Omega / \mathrm{m}$
The resistance is given by $\mathrm{R}=$ Resistance per unit length ÃÂ Length $=\lambda 	imes \mathrm{L}$.
Path 1 (Minor Arc AB):
The shorter curved wire along the circle's circumference. Since $\angle \mathrm{AOB}=90^{\circ}$ (a quarter circle), the length is $\frac{1}{4}$ of the circumference.
$$ \mathrm{L}_1=\frac{1}{4} 	imes 2 \pi \mathrm{r}=\frac{\pi \mathrm{r}}{2} $$
So, the resistance of path 1 is,
$$ \mathrm{R}_1=\lambda\left(\frac{\pi \mathrm{r}}{2}ight) $$
Path 2 (Major Arc AOB):
The longer curved wire along the remaining part of the circle. This is $\frac{3}{4}$ of the circumference.
$$ \mathrm{L}_2=\frac{3}{4} 	imes 2 \pi \mathrm{r}=\frac{3 \pi \mathrm{r}}{2} $$
So, the resistance of path 2 is,
$$ \mathrm{R}_2=\lambda\left(\frac{3 \pi \mathrm{r}}{2}ight) $$
Path 3 (Wire AOB):
The straight wire sections connecting A to the centre O and then O to B . Thus, the two segments AO and OB are in series. It is given that the length of this wire is $2 r$.
$$ \mathrm{L}_3=\mathrm{r}+\mathrm{r}=2 \mathrm{r} $$
So, the resistance of path 3 is,
$$ \mathrm{R}_{34}=\mathrm{R}_3+\mathrm{R}_4=\lambda(2 \mathrm{r}) $$
Since the three paths are in parallel, the equivalent resistance $\left(R_{e q}ight)$ is,
$$ \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_{34}} $$
$$\Rightarrow $$ $\frac{1}{R_{e q}}=\frac{1}{\lambda \pi r / 2}+\frac{1}{3 \lambda \pi r / 2}+\frac{1}{2 \lambda r}$
$$\Rightarrow $$ $\frac{1}{R_{e q}}=\frac{2}{\lambda \pi r}+\frac{2}{3 \lambda \pi r}+\frac{1}{2 \lambda r}$
$$\Rightarrow $$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\lambda \mathrm{r}}\left[\frac{2}{\pi}+\frac{2}{3 \pi}+\frac{1}{2}ight]$
$$\Rightarrow $$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\lambda \mathrm{r}}\left[\frac{12+4+3 \pi}{6 \pi}ight]$
$$\Rightarrow $$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\lambda \mathrm{r}}\left[\frac{16+3 \pi}{6 \pi}ight]$
$$\Rightarrow $$ $$ \mathrm{R}_{\mathrm{eq}}=\lambda \mathrm{r}\left[\frac{6 \pi}{16+3 \pi}ight]=\frac{6 \lambda \pi \mathrm{r}}{3 \pi+16} $$
Therefore, the equivalent resistance between points A and B is $\frac{6 \lambda \pi \mathrm{r}}{3 \pi+16}$.
Hence, the correct option is (B).

Answer

$(\pi+1) 2 r \lambda$

Answer

$\frac{6 \pi \lambda r}{3 \pi+16}$

Answer

$\frac{3 \pi \lambda r}{8}$

Answer

$2 \pi \lambda r$

Current Electricity

A wire of uniform resistance \lambda \Omega / \mathrm{m} is bent into a circle of radius r and another piece of wire with length 2 r is connected between points A and B(\mathrm{AOB}) as shown in figure. The equivalent resistance between points A and B is \_\_\_\_ \Omega.

Choose the correct answer: