Question

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

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

Answer 1

All resistors are equal $R_0=R_1=R_2=R_3=R_4=R_5=R_6=R_7=R_8$

$i_6{R}_6=i_7(R_7+R_8)$ $(R_7=R_8=R_6=k)$

$i_6{R}_6=1(R_7+R_8)$ $(given{i}_7=1mA)$

$i_{6}k=2k$

Current in $R_6=i_6=2mA$

Apply Kirchoff junction law at A

$i_5=i_6+i_7{i}_5=2+1i_5=3mA$

$R_7$ and $R_8$ are in series and their equivalent is parallel with $R_6$ so equivalent circuit becomes :

$(R_0=R_1=R_2=R_3=R_4=R_5=R)$

$i_4{R}_4=i_5(R_5+\frac{2R}{3})i_4\times R=i_5(R+\frac{2R}{3})i_4\times R=3\times \frac{5R}{3}{i}_4=5A$

Apply Kirchoff Junction law at B

$i_3=i_4+i_5=5+3=8mA$

$R_s$ and $\frac{2R}{3}$ are in series and their equivalent is in parallel with $R_4$ , so equivalent circuit becomes :

$i_2{R}_2=i_3(R_3+\frac{5R}{8})$

we know that $(R_2=R_3=R)$

$i_{2}R=8(R+\frac{5R}{8})i_2=13mA$

Apply Kirchoff Junction law at C

$i_1=i_2+i_3{i}_1=21mA$

$R_3\text{and}\frac{5R}{8}$ are in series and their equivalent is parallel with$R_2$,so equivalent circuit become

$i_0{R}_0=i_1(R+\frac{13R}{21})i_{0}R=21\left(\frac{34R}{21}\right)i_0=34mA$

Total current $=i_0+i$

$=34+21=55mA$

Potential difference across $PQ=i_{0}R$

$=34\times 10\times 1\times 1034V$