Question

If the potential difference $V_A-V_B$ for the circuit shown in the figure is given as $(-\frac{20+x}{9}V)n$. Find $x$.

Answer 1

Let the potential at A be zero volt. Let $i$ be the current from C to A.

This means that potential at C, $V_C=\left(i\right)\left(1\right)=i$ volts

Hence, $V_D=i+1$

$V_B=i+2$

$V_E=i+3$

$V_f=i+4$

$V_G=1$

$V_H=2$

$V_I=3$

$V_J=4$

Current from G to C,$i_{GC}=\frac{V_G-V_C}{R_{GC}}=1-i$

Applying Kirchoff's current law at C gives, $i_{CD}=1-2i$

Also inspecting each of the diagonal branches separately gives,

$i_{HD}=i_{IB}=i_{JE}=1-i$

Applying Kirchoff's law on all of points D, B, E gives $i_{EF}=4-8i$

Considering branch FJ gives, $i_{FJ}=\frac{V_F-V_J}{R_{RJ}}=i$

But EF and FJ are in series, hence $4-8i=i$

Hence $i=\frac{4}{9}$

Thus we get $V_A-V_B=-(i+2)=\frac{-22}{9}$volts