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Q.
What current will flow through the $2\, k \Omega$ resistor in the circuit shown in the figure?
Current Electricity
Solution:
In the given circuit $4\, k \Omega$ and $2\, k \Omega$ are in series.
This combination is in parallel with $3\, k \Omega$ resistance. Their effective resistance is
$=\frac{(4+2) \times 3}{(4+2)+3}=2\, k \Omega$
This $2\, k \Omega$ resistance is in series with $6\, k \Omega$.
Thus, the equivalent resistance of the circuit is
$R_{ eq }=(2\, k \Omega)+(6\, k \Omega)=8\, k \Omega$
Current in the circuit is
$I=\frac{72\, V }{8 \times 10^{3}\, \Omega}=9 \times 10^{-3} A =9\, mA$
Current through $2\, k \Omega$ resistor is
$I_{1}=I \times \frac{1}{3}=\frac{9\, mA }{3}=3\, mA$
