Question

Consider the circuit shown below where all resistors are $1k\Omega$.

If a current of magnitude $1mA$ flows through the resistor marked $X$, what is the potential difference measured between points $P$ and $Q$?

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

Answer 1

Answer: (D)

Let current through resistor $X$ is be $i$.

Now, we consider section $dd$,

Equating potential across $dd$, we get
$i_{2}R=i_1(2R)$
or $i_2=2i_1$
Hence, current $i$ is
$i_3=i_1+i_2=i_1+2i_1=3i_1$
Now, we consider section $cc$,

Equating potentials, we get
$i_4=R=i_3(R+\frac{2}{3}R)$
$i_4=i_3(\frac{5}{3})$
So, current, $i_5=i_4+i_3$
$=\frac{5}{3}{i}_3+i_3=\frac{8}{3}{i}_3$
$=\frac{8}{3}(3i_1)=8i_1$
Similarly, across section $bb$,

$\Rightarrow R{i}_6=(\frac{13}{8}R)i_5$ or $i_6=\frac{13}{8}{i}_5$
So, current,
$i_7=\frac{13}{8}{i}_5+i_5$
$=\frac{21}{8}{i}_5=\frac{21}{8}(8i_1)=21i_1$
Now, for section $aa$, we have

$R{i}_8=i_7(\frac{13}{21}+1)R$
$R{i}_8=i_7(\frac{34}{21}R)$
$\Rightarrow i_8=i_7\times \frac{34}{21}$
Hence, current $i$ is
$i=i_7+i_8$
$=i_7+\frac{34}{21}{i}_7=\frac{55}{21}{i}_7$
$=\frac{55}{21}\times 21i_1$
$=55i_1=55\times 10^{-3}A$
$(\because i_1=1mA$, given)
Total resistance across $PQ$ is
$R_{eq}=\frac{34}{55}k\Omega$
$=\frac{34\times 1000}{55}\Omega$
So, potential drop across, $PQ$
$=i{R}_{eq}=55\times 10^{-3}\times \frac{34}{55}\times 10^{3}V$
$=34V$