In the following circuit diagram, $E = 4\,V, r = 1\,\Omega$ and $R = 45\, \Omega$, then reading of the ammeter $A$ will be
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Using Nodal analysis
All resistances are in parallel
Equivalent circuit
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}$
$\mathrm{I}=\frac{\mathrm{V}}{\frac{45}{3}+1}$
$=\frac{4}{16} \Rightarrow \frac{1}{4}$
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