A current of $2.0$ ampere passes through a cell of $e.m.f$. $1.5\, volts$ having internal resistance of $0.15\, ohm$. The potential difference measured, in $volts$, across both the ends of the cell will be
A$1.35$
B$1.5$
C$1$
D$1.2$
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D$1.2$
d (d) $V = E - ir = 1.5 - 2 \times 0.15 = 1.20\,Volt$.
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