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

# Battery charger bridge: average current, efficiency

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

- Asked in: 2081 Baishakh, 8 marks
- Video: [Power Electronics | Lecture - 4A | Single-Phase Full-Wave Diode Rectifiers - Part-1](https://www.youtube.com/watch?v=Dz6SvQzfWIA) — Engineering Devotion
- 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-avg-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

For the circuit shown below, calculate i) Average value of charging current. ii) Efficiency of the charging system. [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

For the 30 V rms diode bridge charging a 24 V battery through 5 ohm, average charging current is 1.49 A. Battery power is 35.80 W, resistor loss 21.84 W, so charging efficiency is 62.1%.

## Answer

**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 $E$.

```text
 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 $= 62.1\%$.

## Check yourself

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

1. In a diode bridge charging a 24 V battery through R, when does a diode pair conduct?
   A. Throughout every half cycle
   B. Only while the instantaneous supply voltage exceeds the battery emf E
   C. Only while the supply voltage is below E
   D. Only after the battery is fully charged
   Answer: B. The battery reverse biases the diodes until |vs| exceeds E, so conduction occurs only between theta1 and pi - theta1. (Key point: Conduction condition)

2. What is the angle at which conduction starts, and its value here (Vm = 42.43 V, E = 24 V)?
   A. theta1 = sin^-1(E/Vm) = 34.45°
   B. theta1 = cos^-1(E/Vm) = 55.55°
   C. theta1 = sin^-1(Vm/E), not possible
   D. theta1 = 0° since diodes need no trigger
   Answer: A. Conduction begins when Vm sin(theta1) = E, giving theta1 = sin^-1(24/42.43) = 34.45°, ending at pi - theta1. (Key point: Conduction start angle)

3. Which expression and result give the average charging current?
   A. Iavg = (Vm - E)/R = 3.69 A
   B. Iavg = (1/(pi R))[2Vm cos(theta1) - E(pi - 2 theta1)] = 1.49 A
   C. Iavg = 2Vm/(pi R) = 5.4 A
   D. Iavg = (Vs - E)/R = 1.2 A
   Answer: B. Average of (Vm sin wt - E)/R over theta1 to pi - theta1 gives the bracketed formula, which evaluates to about 1.49 A. (Key point: Average charging current)

4. How is the charging efficiency obtained in this answer?
   A. Battery power / (battery power + resistor loss) = 35.80/57.64 = 62.1%
   B. Eb / Vs = 24/30 = 80%
   C. Iavg^2 R / (E Iavg) = 61%
   D. Output power / input VA with PF = 1
   Answer: A. Useful power is E*Iavg to the battery; the loss is Irms^2 R in the resistor, so efficiency is about 62.1%. (Key point: Efficiency definition)

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