---
title: "Battery charger bridge 15 V: limit current to 0.5 A"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:battery-bridge-15v-extra-resistance"
source: https://www.bctengineer.com/q/semester_7/power_electronics/battery-bridge-15v-extra-resistance
site: BCT Engineering
---

# Battery charger bridge 15 V: limit current to 0.5 A

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

- Asked in: 2075 Asoj, 8 marks
- 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-15v-extra-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

In figure below shows a full-wave rectifier circuit used to charge a 12 V battery through a 2 Ω resistor. Calculate: i) Average value of charging current ii) Power supplied to the battery iii) Gross output power from rectifier iv) Additional resistance to be connected in series with 2 Ω resistors to limit the average value of charging current to 0.5 amp. [Figure: single-phase diode bridge (four diodes) fed from Vs = 15 V with input current Is; output Vo feeds R = 2 Ω in series with a battery Eb = 12 V; charging current Io]

## Short answer

V_m=21.21 V, α=34.45°. Average charging current Io=1.865 A, battery power=22.38 W, gross rectifier output=36.03 W. Extra series resistance to limit Io to 0.5 A is 5.46 Ω.

## 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: $V_s = 15$ V (rms), $E_b = 12$ V, $R = 2\ \Omega$, ideal diodes.

#### Conduction angle

$$
\begin{aligned}
V_m &= \sqrt2 \times 15 = 21.21\ \text{V} \\
\alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{12}{21.21} = 34.45^\circ = 0.6013\ \text{rad}
\end{aligned}
$$

Current flows for $\pi - 2\alpha = 111.10^\circ$ in each half cycle.

#### i) Average charging current

$$
\begin{aligned}
I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\
&= \frac{1}{\pi \times 2}\left[2(21.21)(0.8246) - 12(1.9391)\right] = \frac{34.986 - 23.269}{6.2832} = 1.865\ \text{A}
\end{aligned}
$$

#### ii) Power supplied to the battery

$$
P_{batt} = E_b I_o = 12 \times 1.865 = 22.38\ \text{W}
$$

#### iii) Gross output power of the rectifier
Gross output = power to battery + loss in $R$. First the rms current:

$$
\begin{aligned}
I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\
&= \frac{715.51 + 209.91 - 839.66}{12.566} = 6.826 \\
I_{rms} &= 2.613\ \text{A}
\end{aligned}
$$

$$
\begin{aligned}
P_{batt} &= E\,I_o = 12 \times 1.865 = 22.38\ \text{W} \\
P_R &= I_{rms}^2R = 6.826 \times 2 = 13.65\ \text{W} \\
P_{gross} &= P_{batt} + P_R = 22.38 + 13.65 = 36.03\ \text{W}
\end{aligned}
$$

#### iv) Extra resistance for $I_o = 0.5$ A
The conduction angle does not depend on $R$, so the bracket $\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] = 11.717$ V stays the same:

$$
\begin{aligned}
R_{total} &= \frac{2V_m\cos\alpha - E(\pi-2\alpha)}{\pi I_o} = \frac{11.717}{\pi \times 0.5} = 7.459\ \Omega \\
R_{extra} &= 7.459 - 2 = 5.459\ \Omega
\end{aligned}
$$

**Answer: (i) $I_o = 1.865$ A, (ii) $P_{batt} = 22.38$ W, (iii) gross output $= 36.03$ W, (iv) extra series resistance $= 5.46\ \Omega$.**

## Check yourself

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

1. In a bridge battery charger, when does a diode pair conduct?
   A. Whenever the supply voltage is positive
   B. Only while the instantaneous supply voltage magnitude exceeds the battery emf E
   C. Only while the supply voltage exceeds E/2
   D. Throughout the whole cycle
   Answer: B. A diode is forward biased only when |vs| is greater than the battery emf; otherwise the battery blocks current. (Key point: Diode conduction condition)

2. With Vm = 21.21 V and E = 12 V, what is the conduction angle per half cycle?
   A. 34.45 degrees
   B. 145.55 degrees
   C. 180 degrees
   D. 111.10 degrees
   Answer: D. alpha = asin(12/21.21) = 34.45 degrees, so conduction lasts pi - 2*alpha = 111.10 degrees. (Key point: Conduction angle)

3. Which expression gives the average charging current of the full-wave circuit?
   A. (1/(2*pi*R))[2Vm cos(alpha) - E(pi-2alpha)]
   B. (1/(pi*R))[2Vm cos(alpha) - E(pi-2alpha)]
   C. (1/(pi*R))[Vm cos(alpha) - E*alpha]
   D. Vm/(pi*R)
   Answer: B. The average over a half cycle is (1/pi) times the integral of (Vm sin wt - E)/R from alpha to pi-alpha. (Key point: Average current expression)

4. Why can the same bracket value 11.717 V be reused to find the extra resistance for Io = 0.5 A?
   A. Because R does not change the conduction angle, which depends only on E and Vm
   B. Because the battery voltage changes with R
   C. Because the diodes conduct for a full half cycle
   D. Because the supply voltage is reduced
   Answer: A. alpha depends on Vm and E only, so only R changes: R_total = 11.717/(pi*0.5) = 7.459 ohm, extra = 5.46 ohm. (Key point: Extra series resistance)

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