Question

The length of uniform wire of resistance $10\Omega$ is bent into a circle and two points at a quarter of circumference apart, are connected to a battery of e.m.f. $2.3V$ and internal resistance $1\Omega$. Then $i_1$ and $i_2$ are

• A. $0.6,0.2A$
• B. $0.2,0.6A$
• C. $0.3,0.1A$
• D. $0.1,0.3A$

Answer 1

The correct option is A $0.6,0.2A$

$\frac{Q}{A}=k$

$R=kl\Rightarrow R=k\left(2\pi r\right)\Rightarrow 10=k\left(2\pi r\right)$

$r=\frac{5}{k\pi }....\left(i\right)$

$l_1=\left(\frac{2\pi r}{4}\right)\Rightarrow \frac{\pi r}{2}\&l_2=2\pi r=\frac{\pi r}{2}$

$=\frac{3}{2}\pi r$

$R_{EF}=\left(\frac{R_1{R}_2}{R_1+R}\right)=\frac{k^2{l}_1{l}_2}{k\left(l_1+l_2\right)}$

$=\frac{k\frac{3}{4}{\left(\pi R\right)}^2}{2\pi R}=\frac{3k\pi r}{8}$

from (i) $R_{EE}=\left(\frac{3}{8}\times 5\right)=\frac{15}{8}$

$I=\frac{E}{R_e+r_1}=\frac{23}{\frac{15}{8}+1}=\left(\frac{24}{23}\right)\Rightarrow 0.8$

$V=R_{EF}\times 0.8$

$V=1.5V$

$I_2=\frac{1.5}{15}\times 2=0.2$

$I_1=0.8-I_2=0.6$