Battery charger bridge: average current, efficiency

Given: single-phase diode bridge, $V_s = 30$ V rms; $R = 5\ \Omega$; battery $E = 24$ V.

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 $= E I_{avg} + I_{rms}^2 R$.

(i) Average charging current

$$\begin{aligned} V_m &= \sqrt{2} \times 30 = 42.43\ \text{V} \\ \theta_1 &= \sin^{-1}\frac{24}{42.43} = 34.45^\circ = 0.6013\ \text{rad}, \qquad \pi - 2\theta_1 = 1.9391\ \text{rad} \\ I_{avg} &= \frac{1}{5\pi}\left[2 \times 42.43 \times 0.8246 - 24 \times 1.9391\right] \\ &= \frac{69.97 - 46.54}{15.708} = 1.492\ \text{A} \end{aligned}$$

(ii) Efficiency of the charging system

$$\begin{aligned} I_{rms}^2 &= \frac{1}{25\pi}\left[(900 + 576)(1.9391) + 900(0.9330) - 4(42.43)(24)(0.8246)\right] = 4.368 \\ I_{rms} &= 2.090\ \text{A} \\ P_b &= E I_{avg} = 24 \times 1.492 = 35.80\ \text{W} \\ P_R &= I_{rms}^2 R = 4.368 \times 5 = 21.84\ \text{W} \\ \eta &= \frac{P_b}{P_b + P_R} = \frac{35.80}{57.64} = 62.1\% \end{aligned}$$

Answer: $I_{avg} = 1.49$ A and charging efficiency $= 62.1\%$.