Voltage and Current Between Points A and B in Circuit

To find the potential difference (voltage) $V_{AB}$ between points $A$ and $B$ and the current $I$ passing through the branch containing the passive resistors between $A$ and $B$, we can model the circuit using Millman's Theorem for parallel branches connected across nodes $A$ and $B$.

1. Identification of Parallel Branches

By observing the circuit diagram:

• Node $A$ corresponds to the entire left vertical bus wire.
• Node $B$ corresponds to the entire right vertical bus wire.

There are 5 parallel branches connected between nodes $A$ and $B$:

1. Branch 1 (Top): Battery of $E_1 = 27\text{ V}$ in series with resistance $R_1 = 3\ \Omega$.
2. Branch 2: Battery of $E_2 = 27\text{ V}$ in series with resistance $R_2 = 3\ \Omega$.
3. Branch 3: Two batteries in series giving $E_3 = 14\text{ V} + 13\text{ V} = 27\text{ V}$ with resistance $R_3 = 3\ \Omega$.
4. Branch 4 (Diagonal Branch): No voltage source ($E_4 = 0\text{ V}$) and two $3\ \Omega$ resistors in series, giving a total resistance of $R_4 = 3\ \Omega + 3\ \Omega = 6\ \Omega$.
5. Branch 5 (Bottom): Battery of $E_5 = 27\text{ V}$ in series with resistance $R_5 = 3\ \Omega$.

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2. Calculation of Voltage $V_{AB}$

According to Millman's Theorem, the potential difference $V_{AB} = V_A - V_B$ across two common nodes is given by:

$V_{AB} = \frac{\sum_{k=1}^{n} \frac{E_k}{R_k}}{\sum_{k=1}^{n} \frac{1}{R_k}}$

Calculating the numerator $\sum \frac{E_k}{R_k}$: $\frac{E_1}{R_1} = \frac{27}{3} = 9\text{ A}$ $\frac{E_2}{R_2} = \frac{27}{3} = 9\text{ A}$ $\frac{E_3}{R_3} = \frac{27}{3} = 9\text{ A}$ $\frac{E_4}{R_4} = \frac{0}{6} = 0\text{ A}$ $\frac{E_5}{R_5} = \frac{27}{3} = 9\text{ A}$

$\sum_{k=1}^{5} \frac{E_k}{R_k} = 9 + 9 + 9 + 0 + 9 = 36\text{ A}$

Calculating the denominator $\sum \frac{1}{R_k}$: $\sum_{k=1}^{5} \frac{1}{R_k} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = 4\left(\frac{1}{3}\right) + \frac{1}{6} = \frac{4}{3} + \frac{1}{6} = \frac{9}{6} = \frac{3}{2}\ \Omega^{-1}$

Substituting these values back into Millman's formula: $V_{AB} = \frac{36}{\frac{3}{2}} = 24\text{ V}$

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3. Calculation of Current $I$

execution The current $I$ flowing through the diagonal branch containing the two $3\ \Omega$ resistors (total resistance $R_4 = 6\ \Omega$) is given by Ohm's Law:

$I = \frac{V_{AB}}{R_4} = \frac{24\text{ V}}{6\ \Omega} = 4\text{ A}$

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Conclusion

• Voltage ($V_{AB}$): $24\text{ V}$
• Current ($I$): $4\text{ A}$

Thus, the correct option is B.