Question

This question relates to the electric circuit above.
At which point is the voltage halved?

• A. A
• B. B
• C. C
• D. D
• E. E

Answer 1

The correct option is D D

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

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

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

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 $I=12/R=12/12=1A$

now total voltage of circuit is $12V$

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

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

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

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

therefore it is clear that at point $D$ voltage becomes $6V$ i.e. dropped to half.