Correct Answer - Option 4 : P / 4

**Concept:**

Power transfer capability of the transmission line is the indication of how much power transfers can be increased without compromising system security.

\(P\propto \frac{{{V}_{S}}\cdot {{V}_{R}}}{{{X}_{S}}}\)

\(or~{{P}_{max}}=\frac{V_{S}^{2}}{{{X}_{S}}}\)

It can be increased either by increasing voltage or by decreasing the reactance of the line.

**Calculation:**

It is given that X_{S} = constant

\({{P}_{max}}=\frac{V_{S}^{2}}{{{X}_{S}}}~or~P\propto V_{S}^{2}\)

At V_{S} = 800 V, P_{m1} ∝ (800)^{2}

And at V_{S} = 400 V, P_{m2} ∝ (400)^{2}

\(\Rightarrow \frac{{{P}_{m1}}}{{{P}_{m2}}}=\frac{{{\left( 800 \right)}^{2}}}{{{\left( 400 \right)}^{2}}}=4\)

⇒ P_{2} = P_{1} / 4

And power transfer capability at V_{S} = 800 V is P.

⇒ Power transfer capability at V_{S} = 400 V

\(\Rightarrow {{P}_{2}}=\frac{P}{4}\)