The analysis of the full-bridge inverter with resistive-load can be carried-out on similar lines of half-bridge inverter with resistive-load. Hence all equations of half-bridge are valid with $E_{\mathrm{dc}} / 2$ replaced by $E_{\mathrm{dc}}$ .

(i) RMS output voltage, $E_{0(\mathrm{rms})}=E_{\mathrm{dc}}$

(ii) Fourier series, $e_{0(\omega t)}=\sum_{n=1,3,5, \ldots}^{\infty} \frac{4 E_{\mathrm{dc}}}{n \pi} \sin (n \omega t)$

(iii) Fundamental output voltage, $E_{0(\text { fund })}=\frac{2 \sqrt{2}}{\pi} \cdot E_{\mathrm{dc}}$

(iv) $n^{\text { th }}$ harmonic voltage: $E_{0}(n)=\frac{E_{0 \text { (fund) }}}{n}$

(v) Transistor (switch) ratings: $V_{\mathrm{CE}_{0}} \geq E_{\mathrm{dc}}, I_{T(\mathrm{av})}=\frac{E_{\mathrm{dc}}}{2 R}$

$$I_{T(\mathrm{rms})}=\frac{E_{\mathrm{dc}}}{\sqrt{2} R}, I_{T(\text { peak })}=\frac{E_{\mathrm{dc}}}{R}$$