In the given figure an ammeter $A$ consists of a $240 \Omega$ coil connected in parallel to a $10 \Omega$ shunt. The reading of the ammeter is . . . . . . $\mathrm{mA}$.
JEE MAIN 2024, Diffcult
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$\text { Req }$
$=140.4+\frac{240 \times 10}{240+10}$
$\text { Req }=140.4+\frac{2400}{250}$
$\text { Req. }=150 \Omega$
$\therefore$ Current in ammeter$ =\frac{24}{150}$
$=160 \mathrm{~mA}$
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