---
title: "Battery charging from 230 V full-wave rectifier"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:battery-230v-200v-emf"
source: https://www.bctengineer.com/q/semester_7/power_electronics/battery-230v-200v-emf
site: BCT Engineering
---

# Battery charging from 230 V full-wave rectifier

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

- Asked in: 2079 Baishakh, 8 marks; 2074 Chaitra, 8 marks (asked 2 times)
- 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-230v-200v-emf
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

Single phase full wave rectifier charges a battery from a single phase supply of 230 V, 50 Hz. The battery has internal emf of 200 volts and its internal resistance is 0.5 ohms. Calculate: (i) Average value of charging current (ii) Power supplied to the battery (iii) Gross power output from the rectifier (iv) Additional resistance to be connected in series to reduce the charging current by 30%.

## Short answer

For the 230 V, 50 Hz full-wave rectifier charging a 200 V battery through 0.5 ohm, conduction starts at sin^-1(200/325.27). Average charging current is 95.24 A, battery power 19.05 kW, rectifier gross output 28.54 kW. To reduce current by 30%, add 0.214 ohm series resistance.

## Answer

**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 $E$.

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

#### 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 $= 0.214\ \Omega$.

## Check yourself

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

1. When does the diode rectifier conduct while charging the battery of emf E?
   A. Throughout the whole half cycle
   B. Only when the instantaneous supply voltage exceeds E, from theta1 = asin(E/Vm) to pi - theta1
   C. Only at the supply peak
   D. From 0 to theta1
   Answer: B. Current flows only when Vm sin(wt) > E, and then i = (Vm sin wt - E)/R. (Key point: Conduction condition)

2. What is the average charging current for 230 V, E = 200 V, R = 0.5 ohm?
   A. 66.67 A
   B. 137.75 A
   C. 95.24 A
   D. 28.5 A
   Answer: C. I_avg = [2 x 325.27 x cos 37.94 - 200 x 1.8171]/(pi x 0.5) = (513.03 - 363.43)/1.5708 = 95.24 A. (Key point: Average charging current)

3. To cut the average charging current by 30%, the added series resistance is about:
   A. 0.5 ohm
   B. 0.15 ohm
   C. 0.35 ohm
   D. 0.214 ohm
   Answer: D. Theta1 is unchanged, so I_avg is proportional to 1/R; R_total = 0.5/0.7 = 0.7143, so R_add = 0.2143 ohm. (Key point: Additional series resistance)

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