Question

ABCD is a cyclic quadrilateral having AC as the diameter of the circle around it All the sides of this quadrilateral including diameter of the circle are formed by the wires having resistance 4 $\Omega$/cm If sides AD and AB are respectively 4 cm and 3 cm and radius of the circle is 2.5 cm then the resistance of the arrangement between B and D will be :

• A. $14\Omega$
• B. $56\Omega$
• C. $\frac{236}{17}\Omega$
• D. None of these

Answer 1

The correct option is C $\frac{236}{17}\Omega$

We know that diameter of a circle subtends right angles at the circumference. Hence $\angle ADC$ and $\angle ABC$ are ${90}^{\circ }$ each.

So $CD=3cm$

$BC=4cm$

Hence the circuit can be redrawn as shown in the second figure, with resistances marked by multiplying length by resistance per unit length.

The currents are marked as shown.

The potential drop across B and D =$12i_1+16i_2$

Considering the loop ABC and applying Kirchoff's Voltage Law,

$12i_1+20(i_1-i_2)$

$⟹i_{=}\frac{9}{8}{i}_2$

Using this in the above equation,

Potential drop=$12\left(\frac{9}{8}{i}_2\right)+16i_2=R_{eq}(i_1+i_2)$

$⟹R_{eq}=\frac{236}{17}\Omega$