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Q. The current in the following circuit isPhysics Question Image

AIPMTAIPMT 1997Current Electricity

Solution:

Applied voltage $( V )=2 V$ and resistances $=3 \Omega, 3 \Omega, 3 \Omega$.
From the given circuit, we find that two resistances are in series and third resistance is in parallel. Therefore equivalent resistance for series resistances $=3+3=6 \Omega$.
Now it is connected parallel with $3 \Omega$ resistance.
Therefore $\frac{1}{R}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2} $ or $ R=2 \Omega \text {. }$
And current flowing in the circuit (l)
$=\frac{V}{R}=\frac{2}{2}=1 A $