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A cell of EMF $1.5$ V and internal resistance $0.5\ \Omega$ delivers $1$ A to a bulb. Terminal voltage is:
A$2$ V, summing EMF and internal drop together
B$0.5$ V, equal to the internal drop alone
C$1.5$ V, the same as EMF here always
D$1$ V, since $V = \varepsilon - Ir = 1.5 - 0.5$
Answer & Solution
Correct answer: D. $1$ V, since $V = \varepsilon - Ir = 1.5 - 0.5$
$V = \varepsilon - Ir = 1.5 - 1\cdot 0.5 = 1.0$ V.
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