---
title: "Battery charger bridge 15 V: currents, power, efficiency"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:battery-bridge-15v-efficiency"
source: https://www.bctengineer.com/q/semester_7/power_electronics/battery-bridge-15v-efficiency
site: BCT Engineering
---

# Battery charger bridge 15 V: currents, power, efficiency

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

- Asked in: 2076 Chaitra, 8 marks; 2072 Kartik, 8 marks (asked 2 times)
- 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-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

Figure shows a full wave rectifier circuit used to charge a 12V battery through a 2 ohm resistor. (i) Draw the waveform of output voltage V0 and charging current i0. (ii) Calculate the average and rms value of charging current. (iii) Power supplied to the battery. (iv) Power output from the rectifier. (v) Efficiency of the charging system. [Figure: single-phase diode bridge (four diodes) fed from Vs = 15 V (rms), f = 50 Hz; output Vo feeds R = 2 Ω in series with a battery Eb = 12 V; charging current io]

## Short answer

For the 15 V rms bridge charging a 12 V battery through 2 ohm, conduction occurs only when rectified voltage exceeds 12 V. Average charging current is 1.865 A, rms current 2.613 A, battery power 22.38 W, rectifier output 36.03 W. Charging efficiency is 62.1%.

## Answer

**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 $E$.

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

#### (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, $\eta = 62.1\%$.

## Check yourself

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

1. For Vs = 15 V rms and a 12 V battery, conduction starts at theta1 equal to:
   A. asin(12/21.21) = 34.45 degrees
   B. asin(12/15) = 53.1 degrees
   C. 0 degrees
   D. 90 degrees
   Answer: A. Vm = sqrt(2) x 15 = 21.21 V and theta1 = asin(E/Vm) = 34.45 degrees. (Key point: Conduction start angle)

2. How is the power supplied to the battery calculated and what is it?
   A. I_rms^2 R = 13.65 W
   B. E x I_avg = 12 x 1.865 = 22.38 W
   C. Vs x I_avg = 15 x 1.865 W
   D. E x I_rms = 12 x 2.613 W
   Answer: B. The battery stores only the average-current power, E times I_avg; the I_rms^2 R term is lost in the resistor. (Key point: Power to battery)

3. What is the efficiency of the charging system?
   A. 100 percent, since the diodes are ideal
   B. 37.9 percent
   C. 62.1 percent
   D. 82.5 percent
   Answer: C. Rectifier output is 22.38 + 2.613^2 x 2 = 36.03 W, so efficiency = 22.38/36.03 = 62.1%. (Key point: Efficiency)

## 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×

## Related in Power Electronics

- [Battery charger bridge 50 V: power and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-50v-efficiency) — Power Electronics, chapter 2
- [Battery charger bridge 15 V: limit current to 0.5 A](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-15v-extra-resistance) — Power Electronics, chapter 2
- [Battery charger bridge 30 V: conduction, power](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-conduction-power) — Power Electronics, chapter 2, asked 2×
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- [Battery charger bridge: average current, efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-avg-efficiency) — Power Electronics, chapter 2
- [Half-wave diode battery charger, 10 A](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahalf-wave-battery-charger-24v) — Power Electronics, chapter 2

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