Battery charger bridge 15 V: currents, power, efficiency

Given: single-phase diode bridge, $V_s = 15$ V rms, 50 Hz; $R = 2\ \Omega$; battery $E = 12$ V.

(i) Waveforms

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

(ii) Average and rms charging current

$$\begin{aligned} V_m &= \sqrt{2} \times 15 = 21.21\ \text{V} \\ \theta_1 &= \sin^{-1}\frac{12}{21.21} = 34.45^\circ = 0.6013\ \text{rad}, \qquad \pi - 2\theta_1 = 1.9391\ \text{rad} \\ I_{avg} &= \frac{1}{2\pi}\left[2 \times 21.21 \times 0.8246 - 12 \times 1.9391\right] = \frac{34.99 - 23.27}{6.283} = 1.865\ \text{A} \end{aligned}$$

With $\sin 2\theta_1 = 0.9330$:

$$I_{rms}^2 = \frac{1}{4\pi}\left[(225 + 144)(1.9391) + 225(0.9330) - 4(21.21)(12)(0.8246)\right] = 6.826$$ $$I_{rms} = 2.613\ \text{A}$$

(iii) Power supplied to the battery

$$P_b = E I_{avg} = 12 \times 1.865 = 22.38\ \text{W}$$

(iv) Power output of the rectifier

$$P_o = P_b + I_{rms}^2 R = 22.38 + 2.613^2 \times 2 = 22.38 + 13.65 = 36.03\ \text{W}$$

(v) Efficiency of charging

$$\eta = \frac{P_b}{P_o} = \frac{22.38}{36.03} = 62.1\%$$

Answer: $I_{avg} = 1.865$ A, $I_{rms} = 2.613$ A, $P_b = 22.38$ W, $P_o = 36.03$ W, $\eta = 62.1\%$.