Question

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

If a current of magnitude $1 \,mA$ 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
$i_2R = i_1 (2R)$
or $i_2 = 2 i_1$
Hence, current $i$ is
$i_3 = i_1 + i_2 = i_1 + 2 i_1 = 3 i_1$
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) = 8\,i_1$
Similarly, across section $bb$,

$\Rightarrow Ri_6 = (\frac{13}{8}R) i_5$ 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) = 21 \,i_1$
Now, for section $aa$, we have

$Ri_8 = i_7 (\frac{13}{21} + 1) R$
$Ri_8 = i_7(\frac{34}{21}R)$
$\Rightarrow i_8 = i_7 \times \frac{34}{21}$
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$
$(\because i_1 = 1\,mA$, given)
Total resistance across $PQ$ is
$R_{eq} = \frac{34}{55} \,k\,\Omega$
$ = \frac{34 \times 1000}{55} \Omega$
So, potential drop across, $PQ$
$= i\, R_{eq} = 55 \times 10^{-3} \times \frac{34}{55} \times 10^3 \,V$
$ = 34\,V$