Battery charging from 230 V full-wave rectifier

Given: single-phase full-wave diode rectifier, $V_s = 230$ V, 50 Hz; battery $E = 200$ V, internal resistance $R = 0.5\ \Omega$.

In a diode rectifier charging a battery of emf $E$ through resistance $R$, a diode pair conducts only when the instantaneous supply voltage exceeds $E$.

 v  ^      .--.          .--.
    |    .'    '.      .'    '.  vs (rectified)
  E |---+--------+----+--------+--- battery emf
    |  /|        |\  /|        |
    | / |  io    | \/ |        |
 ---+---+--------+----+--------+--> wt
       t1     pi-t1  pi+t1
 io  ^   .--.          .--.
     |  /    \        /    \
 ----+-+------+------+------+--> wt

Conduction starts at $\theta_1 = \sin^{-1}(E/V_m)$ and ends at $\pi - \theta_1$ in every half cycle. During conduction $i_o = \frac{V_m\sin\omega t - E}{R}$.

$$\begin{aligned} I_{avg} &= \frac{1}{\pi R}\left[2V_m\cos\theta_1 - E(\pi - 2\theta_1)\right] \\ I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2} + E^2\right)(\pi - 2\theta_1) + \frac{V_m^2}{2}\sin 2\theta_1 - 4V_mE\cos\theta_1\right] \end{aligned}$$

Power to the battery $= E I_{avg}$; loss in the resistor $= I_{rms}^2 R$; rectifier output $= E I_{avg} + I_{rms}^2 R$.

Conduction angle

$$\begin{aligned} V_m &= \sqrt{2} \times 230 = 325.27\ \text{V} \\ \theta_1 &= \sin^{-1}\frac{200}{325.27} = 37.94^\circ = 0.6622\ \text{rad}, \qquad \pi - 2\theta_1 = 1.8171\ \text{rad} \end{aligned}$$

(i) Average charging current

$$\begin{aligned} I_{avg} &= \frac{1}{\pi \times 0.5}\left[2 \times 325.27 \times \cos 37.94^\circ - 200 \times 1.8171\right] \\ &= \frac{513.03 - 363.43}{1.5708} = 95.24\ \text{A} \end{aligned}$$

(ii) Power supplied to the battery

$$P_b = E I_{avg} = 200 \times 95.24 = 19048\ \text{W} \approx 19.05\ \text{kW}$$

(iii) Gross power output of the rectifier

Using the rms formula with $\sin 2\theta_1 = 0.9698$, $\cos\theta_1 = 0.7886$:

$$I_{rms} = 137.75\ \text{A}, \qquad I_{rms}^2 R = 137.75^2 \times 0.5 = 9487.5\ \text{W}$$ $$P_{gross} = P_b + I_{rms}^2 R = 19048 + 9487.5 = 28535.5\ \text{W} \approx 28.54\ \text{kW}$$

(iv) Extra series resistance for 30% less current

$\theta_1$ depends only on $E$ and $V_m$, so $I_{avg} \propto 1/R_{total}$. For $I_{avg}' = 0.7 I_{avg}$:

$$\begin{aligned} R_{total} &= \frac{0.5}{0.7} = 0.7143\ \Omega \\ R_{add} &= 0.7143 - 0.5 = 0.2143\ \Omega \end{aligned}$$

(New $I_{avg} = 0.7 \times 95.24 = 66.67$ A.)

Answer: $I_{avg} = 95.24$ A; $P_b = 19.05$ kW; $P_{gross} = 28.54$ kW; additional resistance $= 0.214\ \Omega$.