In the figure, current through the $3\,\Omega $ resistor is $0.8\, ampere$, then potential drop through $4\,\Omega $ resistor is ........... $V$
AIPMT 1993, Medium
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Current through $6\,\Omega$ resistance in parallel with $3\,\Omega$ resistance $= 0.4\, A$
So total current $=$ $0.8 + 0.4 = 1.2\, A$
Potential drop across $4\,\Omega = 1.2 \times 4 = 4.8\,V$
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