Question

The terminals of a $18V$ battery with an internal resistance of $1.5\Omega$ are connected to a circular wire of resistance $24\Omega$ at two points distant at one quarter of the circumference of a circular wire. The current through the bigger arc of the circle will be

• A. $0.75A$
• B. $1.5A$
• C. $2.25A$
• D. $3A$

Answer 1

The correct option is A $0.75A$

Since the terminal is connected at one quarter distance .

The ratios of length of two points is $\frac{L_1}{L_2}=\frac{3\pi /2}{\pi /2}$

$R=24\Omega$

Since $R\propto L$..so $\frac{R_1}{R_2}=\frac{L_1}{L_1}=3$

$R_1=18\Omega and{R}_2=6\Omega$

Since the potential across $R_1$ and $R_2$ is same ,so they both are parallel.

$\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}$ ; $R_c=4.5\Omega$

Now the internal Resistance is $1.5\Omega$

$R_T=1.5+4.5=6\Omega$

Cuurent, $I=\frac{V}{R_T}=\frac{18}{6}=3A$

Now the current in parallel resistance is distributed in inverse ratio of resistance .

Curren through bigger arc $R_1$

$I_1=\frac{R_2}{R_1+R_2}I=\frac{6}{24}\times 3=0.75A$