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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Current Electricity
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A wire of resistance $10\Omega$ is bent to form a circle. $P$ and $Q$ are two points on it such that the arc $PQ$ subtends a right angle at the center. A $3V$ battery having internal resistance of $ 1\Omega$ is connected across $P$ and Q, the currents in the two parts of the circle are

$\begin {array} {1 1} (A)\;\large\frac{5}{26}A\: and \: \large\frac{15}{26}A & \quad (B)\;\large\frac{4}{25}A\: and \: \large\frac{12}{25}A \\ (C)\;\large\frac{3}{35}A\: and \: \large\frac{9}{25}A & \quad (D)\;\large\frac{6}{23}A\: and \: \large\frac{18}{23}A \end {array}$


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When the wire is bent to form a circle arc PQ subtends 90 degrees at the centre so the resistance of smaller part is 1/4 th of 10 ohms and bigger part is 3/4 of 10 ohms.
The resistance of the smaller arc is $2.5\Omega$ and that of the larger arc is $7.5\Omega$.
They are connected in parallel to each other and in series to the internal resistance.
total resiatance in the circuit is  sum of the resiatances in parallel plus the internal resistance of the cell
 Total resistance = (2.5) (7.5)/ 10  +  1    = 2.875
now total curreent in the circuit = 3/2.875  = 24/23
the current in the two branches of the circle will split  according to the inverse of ratio of the resistance i.e. 1 : 3
i.e. 6/23 and 18/23 
Ans : (D)


answered Feb 10, 2014 by thanvigandhi_1
edited Sep 3, 2014 by thagee.vedartham

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