Question

In the figure shown, self-impedances of the two transmission lines are 1.5j p.u each, and $Z_m = 0.5j$ p.u is the mutual impedance. Bus voltages shown in the figure are in p.u. Given that $\delta > 0$, the maximum steady-state real power that can be transferred in p.u from Bus-1 to Bus-2 is 

• A. $|E||V|$
• B. $\dfrac{|E||V|}{2}$
• C. $2|E||V|$
• D. $\dfrac{3|E||V|}{2}$

Hint:

Coupled parallel transmission lines reduce effective reactance, increasing transfer capability. For maximum power, always substitute $\delta = 90^\circ$.

Answer 1

The Correct Option is A

Step 1: Compute the net transfer reactance.
Each line reactance = $1.5j$ p.u.
Mutual reactance = $0.5j$ p.u.
For two coupled parallel lines, the positive-sequence transfer reactance is: \[ X_{\text{eq}} = X - X_m = 1.5 - 0.5 = 1.0 \text{ p.u} \]

Step 2: Use the standard power transfer formula.
The steady-state real power transferred is: \[ P = \frac{|E||V|}{X_{\text{eq}}}\sin \delta \] Given $\delta > 0$, the maximum power occurs at $\delta = 90^\circ$: \[ P_{\max} = \frac{|E||V|}{X_{\text{eq}}} \]

Step 3: Substitute $X_{\text{eq} = 1$.}
\[ P_{\max} = |E||V| \]

Step 4: Conclusion.
Thus the maximum real power transferred from Bus-1 to Bus-2 is: \[ P_{\max} = |E||V| \]