Question

This question relates to the electric circuit above.
Through which point does half the total current pass?

• A. A
• B. B
• C. C
• D. D
• E. E

Answer 1

Answer: (C)

C

Equivalent resistance between point $B$ and $D$ ,

$R^{\prime \prime }=6+6=12\Omega$

and $d\frac{1}{R^{\prime }}=\frac{1}{12}+\frac{1}{12}$

or $R^{\prime }=6\Omega$

Now equivalent resistance between $B$ and $E$ ,

$R=6+R^{\prime }=6+6=12\Omega$

by Kirchoff's law for closed mesh $ABDEA$ ,

$IR=6+6$

or $I=12/R=12/12=1A$ (that is the total current of circuit)

voltage drop between $B$ and $D$ is ,

$V_{BD}=I{R}^{\prime }=1\times 6=6V$

so current in the resistance of $12\Omega$ ,

$I^{\prime }=V_{BD}/12=6/12=0.5A$

and current (through point $C$) in the resistances of $6\Omega$ ,which are in series(current will be same in both resistances) ,

$I^{\prime \prime }=V_{BD}/6+6=6/12=0.5A$

therefore it is clear that through point $C$ , half of the total current passes.