Considering two parallel short transmission lines of impedance ZA and ZB respectively. Current IA and IB are b

Question

Considering two parallel short transmission lines of impedance ZA and ZB respectively. Current IA and IB are both lagging and sending end voltage is Vs. If the reactance to resistance ratio of both the impedance ZA and ZB are equal then total current I will be

Examoogle · Accepted Answer

Option C: In phase with both $I_A$ and $I_B$

Explanation & Concepts:
When two parallel lines have identical $X/R$ ratios, their impedance angle $	heta = 	an^{-1}(X/R)$ is identical. Consequently, both branch currents $I_A$ and $I_B$ have the same phase angle relative to the sending end voltage $V_s$, causing them to be in phase with each other.

Working Principle: The current in a line is given by $\vec{I} = \frac{\vec{V_s}}{\vec{Z}}$. Since $\vec{Z_A} = |Z_A| \angle 	heta$ and $\vec{Z_B} = |Z_B| \angle 	heta$, both currents lag the voltage $V_s$ by the same angle $	heta$. If two vectors lag a reference vector by the same angle, the angle between the two vectors themselves is zero, meaning they are in phase.

Key Formulas: $	heta_A = 	heta_B = 	an^{-1}\left(\frac{X_A}{R_A}ight) = 	an^{-1}\left(\frac{X_B}{R_B}ight)$ — Equality of impedance angles; $\vec{I} = \vec{I_A} + \vec{I_B} = \vec{V_s}(\frac{1}{Z_A} + \frac{1}{Z_B})$ — Total current calculation

Key Points: Parallel branches with identical impedance angles force branch currents to have identical power factors.; The total current $I$ is simply the vector sum of $I_A$ and $I_B$, which are collinear.; This condition simplifies load sharing calculations in power system analysis.

Advantages: Uniform power factor distribution.; Simplifies parallel line balancing calculations.

Disadvantages: Requires impedance matching of lines.; Not always physically achievable in existing networks.

Applications: Parallel transmission line modeling.; Load flow studies in power grids.

Additional Info: The condition $\frac{X_A}{R_A} = \frac{X_B}{R_B}$ ensures that the reactive power to active power ratio is identical for both branches.. Option A and B are incorrect because the phase shift relative to $V_s$ is fixed for both lines due to the matching impedance angles.

Answer: C is correct — Since both lines have equal $X/R$ ratios, their impedance phase angles are identical, resulting in $I_A$ and $I_B$ having the same phase angle, making them in phase with each other.

Exam Tip: Always check the $X/R$ ratio when dealing with parallel branches; if they are equal, the currents will always be in phase with each other regardless of the magnitude of the individual impedances.

Examoogle · Answer

Option A: Lagging by both $I_A$ and $I_B$

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Option B: Leading by both $I_A$ and $I_B$

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Option D: None of above

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