Question

Consider the circuit shown in the figure where all the resistances are of magnitude $1k\Omega$. If the current in the extreme right resistance $X$ is 1 $mA$, then the potential difference between $A$ and $B$ is:

• A. $34V$
• B. $21V$
• C. $68V$
• D. $55V$

Answer 1

The correct option is A $34V$

$V_{EFB}=1mA\times 2X=2V$

$X_3$ is parallel to the path $EFB$, $V_{X3}=2V$

$I_{X3}=\frac{2}{1k\Omega }=2mA$

Using KCL, $I_{X4}=I_{X3}+I_{X2}=3mA$

Now, $V_{DEB}=V_{DE}+V_{EB}$

$=3mA\times 1k\Omega +2=5V$

$X_5$ is parallel to the path $DEB$, $V_{X5}=5V$

$I_{X5}=\frac{5}{1k\Omega }=5mA$

Using KCL, $I_{X6}=I_{X5}+I_{X4}=8mA$

Now, $V_{CDB}=V_{CD}+V_{DB}$

$=8mA\times 1k\Omega +5=13V$

$X_7$ is parallel to the path $CDB$, $V_{X7}=13V$

$I_{X7}=\frac{13}{1k\Omega }=13mA$

Using KCL,

$I_{X8}=I_{X7}+I_{X6}=21mA$

Now, $V_{ACB}=V_{AC}+V_{CB}$

$=21mA\times 1k\Omega +13=34V$

$X_9$ is parallel to the path $ACB$, $V_{X9}=34V=V_{AB}$