---
title: "Battery charger bridge 30 V: conduction, power"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:battery-bridge-30v-conduction-power"
source: https://www.bctengineer.com/q/semester_7/power_electronics/battery-bridge-30v-conduction-power
site: BCT Engineering
---

# Battery charger bridge 30 V: conduction, power

Power Electronics · Chapter 2: Single phase AC to DC conversion

- Asked in: 2073 Chaitra, 8 marks
- Video: [Single Phase Full Wave Controlled Rectifier with RLE Load | Power Electronics | Lecture 46](https://www.youtube.com/watch?v=4OqH3R5uNJ8) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/battery-bridge-30v-conduction-power
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

Below figure shows a full-wave diode rectifier circuit used to charge a 24V battery through a 5 ohm resistor. Calculate: (i) Conduction period of charging current io. (ii) Average value of charging current io (iii) Power supplied to the battery and gross power output from the rectifier [Figure: single-phase diode bridge (D1, D3 top; D4, D2 bottom) fed from Vs = 30 V (rms); output Vo feeds R = 5 Ω in series with a battery Eb = 24 V; charging current io]

## Short answer

Conduction period per half cycle is 111.10° (6.17 ms) from 34.45° to 145.55°. Average charging current I_o = 1.492 A. Power to battery is 35.80 W; gross rectifier output power is 57.65 W.

## Answer

In a **battery-charging rectifier**, a diode conducts only when the instantaneous supply voltage exceeds the battery emf $E$. In each half cycle current flows from $\alpha$ to $\pi-\alpha$, where $V_m\sin\alpha = E$, and is limited by $R$:

$$
i_o = \frac{V_m\sin\omega t - E}{R}, \qquad \alpha < \omega t < \pi - \alpha
$$

```text
     io     R
 P o--->---/\/\/---+
                   |
                 + |
                 ===  Eb (battery)
                 - |
 N o---------------+
 (P, N = output of the diode bridge)
```

Data: $V_s = 30$ V (rms), $E_b = 24$ V, $R = 5\ \Omega$, ideal diodes. D1, D2 conduct in the positive half cycle and D3, D4 in the negative half cycle, but only while $|v_s| > 24$ V.

```text
 |vs|   ___         ___
       / | \       / | \       --- 24 V level
 -----/--+--\-----/--+--\----
 io      /\          /\
 -------/  \--------/  \-----
      a  180-a  180+a  360-a
```

#### i) Conduction period

$$
\begin{aligned}
V_m &= \sqrt2 \times 30 = 42.43\ \text{V} \\
\alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{24}{42.43} = 34.45^\circ = 0.6013\ \text{rad}
\end{aligned}
$$

$$
\begin{aligned}
\theta_c &= \pi - 2\alpha = 180^\circ - 2(34.45^\circ) = 111.10^\circ = 1.9391\ \text{rad} \\
t_c &= \frac{111.10^\circ}{360^\circ} \times 20\ \text{ms} = 6.17\ \text{ms (in each half cycle)}
\end{aligned}
$$

#### ii) Average charging current

$$
\begin{aligned}
I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\
&= \frac{1}{\pi \times 5}\left[2(42.43)(0.8246) - 24(1.9391)\right] = \frac{69.971 - 46.538}{15.7080} = 1.492\ \text{A}
\end{aligned}
$$

#### iii) Power to battery and gross power output

$$
\begin{aligned}
I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\
&= \frac{2862.06 + 839.66 - 3358.63}{78.540} = 4.368 \\
I_{rms} &= 2.090\ \text{A}
\end{aligned}
$$

$$
\begin{aligned}
P_{batt} &= E\,I_o = 24 \times 1.492 = 35.80\ \text{W} \\
P_R &= I_{rms}^2R = 4.368 \times 5 = 21.84\ \text{W} \\
P_{gross} &= P_{batt} + P_R = 35.80 + 21.84 = 57.65\ \text{W}
\end{aligned}
$$

**Answer: (i) conduction 111.10° (6.17 ms) per half cycle, from 34.45° to 145.55°; (ii) $I_o = 1.492$ A; (iii) $P_{batt} = 35.80$ W, gross output $= 57.65$ W.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For Vs = 30 V rms and E = 24 V, which relation gives the angle where conduction starts?
   A. Vm cos(alpha) = E
   B. Vs sin(alpha) = E
   C. Vm sin(alpha) = E
   D. E sin(alpha) = Vm
   Answer: C. Conduction starts when Vm sin(wt) reaches E, so alpha = asin(24/42.43) = 34.45 degrees. (Key point: Starting angle)

2. What is the conduction period in each half cycle of a 50 Hz supply?
   A. 111.10 degrees, about 6.17 ms
   B. 145.55 degrees, about 8.1 ms
   C. 69 degrees, about 3.8 ms
   D. 180 degrees, 10 ms
   Answer: A. pi - 2*alpha = 111.10 degrees, and 111.10/360 x 20 ms = 6.17 ms. (Key point: Conduction period)

3. What is the average charging current?
   A. 3.0 A
   B. 0.75 A
   C. 2.09 A
   D. 1.492 A
   Answer: D. Io = [2(42.43)(0.8246) - 24(1.9391)]/(5*pi) = 23.43/15.708 = 1.492 A; 2.09 A is the rms current. (Key point: Average current)

4. How is the gross output power of the rectifier obtained?
   A. Power to battery plus I_rms^2 R loss in the resistor
   B. E x Io only
   C. Vs x Io
   D. I_rms^2 R only
   Answer: A. P_gross = 35.80 W + 21.84 W = 57.65 W. (Key point: Gross output power)

## Close variants set in other exams

- [Battery charger bridge 30 V: conduction period](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-conduction) — Power Electronics, chapter 2, asked 4×

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