The reading of the ammeter as per the figure | vedclass

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(B) The circuit consists of three $2\,\Omega$ resistors connected in parallel between points $X$ and $Y$.The equivalent resistance $R_p$ of these thre

Vedclass · Accepted Answer

$\frac{3}{4}\,A$

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(B) The circuit consists of three $2\,\Omega$ resistors connected in parallel between points $X$ and $Y$.The equivalent resistance $R_p$ of these three parallel resistors is given by:$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \implies R_p = \frac{2}{3}\,\Omega$.This parallel combination is in series with an internal resistance of $2\,\Omega$ (associated with the battery branch).Total resistance of the circuit $R_{eq} = 2 + \frac{2}{3} = \frac{6+2}{3} = \frac{8}{3}\,\Omega$.The current $I$ through the ammeter is given by Ohm's law:$I = \frac{V}{R_{eq}} = \frac{2}{8/3} = 2 	imes \frac{3}{8} = \frac{6}{8} = \frac{3}{4}\,A$.

The reading of the ammeter as per the figure shown is:

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