In the figure shown, if the internal resistance of the battery is $1\, ohm$ , the reading of the ammeter will be ................. $\mathrm{A}$
Medium
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$R_{\text {net }}=\frac{6 \times 3}{6+3}+1=3\, \Omega$
$I_{\text {net }}=\frac{9}{3}=3 \mathrm{\,A}$
$T.P.D.$ of cell $=E-I r=9-3 \times 1=6 \,V$
reading of ameter $=\frac{6}{3}=2 \mathrm{\,A}$
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