Secondary voltage, regulation, pf for max regulation

Given: 10 kVA, 2500/500 V, $R_1 = 5.5\ \Omega$, $X_1 = 12\ \Omega$, $R_2 = 0.2\ \Omega$, $X_2 = 0.45\ \Omega$, $V_1 = 2000$ V, load pf $0.8$ lagging at full load.

Refer everything to the secondary ($K = 500/2500 = 0.2$, $K^2 = 0.04$):

$$\begin{aligned} R_{02} &= R_2 + K^2R_1 = 0.2 + 0.04\times5.5 = 0.42\ \Omega \\ X_{02} &= X_2 + K^2X_1 = 0.45 + 0.04\times12 = 0.93\ \Omega \\ Z_{02} &= \sqrt{0.42^2 + 0.93^2} = 1.0204\ \Omega \end{aligned}$$

Full-load current and no-load secondary voltage

$$I_2 = \frac{10\,000}{500} = 20\ \text{A},\qquad E_2 = KV_1 = 0.2\times2000 = 400\ \text{V}$$

Since the exciting current is neglected, the secondary no-load voltage is 400 V with the 2000 V supply.

Secondary terminal voltage (with $\cos\phi = 0.8$, $\sin\phi = 0.6$). Using $V_2$ as reference:

$$\begin{aligned} E_2^2 &= (V_2 + I_2R_{02}\cos\phi + I_2X_{02}\sin\phi)^2 + (I_2X_{02}\cos\phi - I_2R_{02}\sin\phi)^2 \\ I_2R_{02}\cos\phi + I_2X_{02}\sin\phi &= 20(0.336 + 0.558) = 17.88\ \text{V} \\ I_2X_{02}\cos\phi - I_2R_{02}\sin\phi &= 20(0.744 - 0.252) = 9.84\ \text{V} \\ 400^2 &= (V_2 + 17.88)^2 + 9.84^2 \Rightarrow V_2 = 382.0\ \text{V} \end{aligned}$$

Percentage regulation

$$\text{VR} = \frac{400 - 382.0}{400}\times100 = 4.50\%$$

(The approximate formula gives $\dfrac{17.88}{400}\times100 = 4.47\%$.)

Power factor for maximum regulation

Regulation is greatest when $\tan\phi = X_{02}/R_{02}$, i.e.

$$\cos\phi = \frac{R_{02}}{Z_{02}} = \frac{0.42}{1.0204} = 0.4116\ \text{lagging}$$

The maximum regulation is $\dfrac{I_2Z_{02}}{E_2}\times100 = \dfrac{20\times1.0204}{400}\times100 = 5.10\%$.

Answer: $V_2 \approx 382$ V; regulation $\approx 4.5\%$; pf for maximum regulation $= 0.412$ lagging (VR$_{max} = 5.1\%$)