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