The terminals of an 18 V battery with internal resistance 1.5 Ω are connected to a circular coil of resistance 24 Ω at two points distant one quarter of the circumference of the coil as shown in fig. The current flowing through the bigger part will be

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The terminals of an 18 V battery with internal resistance 1.5 $Ω$ are connected to a circular coil of resistance 24 $Ω$ at two points distant one quarter of the circumference of the coil as shown in fig. The current flowing through the bigger part will be

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0.75A

1.5A

0.5 A

1.0A

answer is B.

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Detailed Solution

See fig. The resistance between points, A and C is given by
$R_{AC} = \frac{18 \times 6}{18 + 6} = 4 \cdot 5 Ω$
Total resistance of the circuit $= 4 \cdot 5 Ω + 1 \cdot 5 Ω = 6 Ω$
Total current $= i = E / R = 18 / 6 = 3 A$
Now $V_{ABC} = V_{AC}$
$\begin{array}{l} i \times 18 = (3 - i) 6 or 24 i = 18 \\ ∴ i = \frac{18}{24} = 0 \cdot 75 A \end{array}$