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

Options

A. 0.225 A
B. 3A
C. 0.75 A
D. 1.5 A

Explanation

ACB and ADB can be replaced by `(1)/(18)+(1)/(6)rArr(4)/(18)=(1)/(R )=4.5Omega`. If internal resistance had been `24Omega`
`(18V)/((24+4.5))=0.84A`.
`therefore` Current in the bigger circle `=(0.84)/(4)=0.21A`. Option is not given. An internal resistance of `24Omega` for a battery appears to be very high.