Per unit resistance drop of core type transformer

Per-unit resistance drop is the full-load resistive voltage drop expressed as a fraction of rated voltage, $\varepsilon_r = I_pR_p/V_p$. It can be found from winding dimensions as below.

Notation

• $T_p$, $T_s$ = primary and secondary turns; $I_p$, $I_s$ = currents
• $L_{mtp}$, $L_{mts}$ = mean lengths of turn
• $a_p$, $a_s$ = conductor areas; $\delta_p$, $\delta_s$ = current densities
• $\rho$ = resistivity of copper; $E_t$ = volts per turn

Derivation

Resistance of each winding:

$$r_p = \frac{\rho T_pL_{mtp}}{a_p},\qquad r_s = \frac{\rho T_sL_{mts}}{a_s}$$

Total resistance referred to primary:

$$R_p = r_p + r_s\left(\frac{T_p}{T_s}\right)^2$$

Per-unit resistance:

$$\varepsilon_r = \frac{I_pR_p}{V_p} = \frac{I_pr_p}{V_p} + \frac{I_pr_s(T_p/T_s)^2}{V_p}$$

Since $I_pT_p = I_sT_s$ and $V_p/T_p = V_s/T_s = E_t$, the second term equals $I_sr_s/V_s$. So

$$\varepsilon_r = \frac{I_pr_p}{V_p} + \frac{I_sr_s}{V_s}$$

Now $I_pr_p = I_p\dfrac{\rho T_pL_{mtp}}{a_p} = \rho\,\delta_p\,T_pL_{mtp}$ (as $I_p/a_p = \delta_p$), and $V_p = E_tT_p$:

$$\frac{I_pr_p}{V_p} = \frac{\rho\,\delta_pL_{mtp}}{E_t},\qquad \frac{I_sr_s}{V_s} = \frac{\rho\,\delta_sL_{mts}}{E_t}$$

Therefore

$$\varepsilon_r = \frac{\rho\left(\delta_pL_{mtp} + \delta_sL_{mts}\right)}{E_t}$$

If both windings have the same current density $\delta$ and a common mean turn $L_{mt}$:

$$\varepsilon_r = \frac{2\rho\,\delta\,L_{mt}}{E_t}$$

Other forms

Multiplying the numerator and denominator by $I_p$:

$$\varepsilon_r = \frac{I_p^2R_p}{V_pI_p} = \frac{\text{full-load copper loss (per phase)}}{\text{VA rating (per phase)}}$$

Example

With $\rho$ = 0.021 Ω-mm²/m, $\delta$ = 2.5 A/mm², $L_{mt}$ = 0.9 m and $E_t$ = 7.5 V:

$$\varepsilon_r = \frac{2\times0.021\times2.5\times0.9}{7.5} = 0.0126\ \text{p.u.} = 1.26\%$$

Conclusions

• $\varepsilon_r$ falls if $E_t$ is high (fewer turns) or $\delta$ is low.
• A short mean turn (stepped core, compact windings) reduces both copper loss and $\varepsilon_r$.