Given:A 55 VDC parking lot light (on only at night) consumes 24 W when on• It is powered by a 55 VDC battery which is recharged by a solar panelThe 111 W solar panel only generates 12 VDC, so an inverter is used to boost the voltage to 55 VDC torecharge the battery• The inverter is 90% efficient (output power is 90% of input power)• After four particularly cloudy days (were the solar panel was unable to recharge the battery), the solar panel isfinally able to recharge the half discharged battery to full in 4.2 hoursWhat is the rated capacity of the 55 VDC battery (in Ah)? Answer tolerance is 5%.

Answered: Given: • A 55 VDC parking lot light (on…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Given:
A 55 VDC parking lot light (on only at night) consumes 24 W when on
• It is powered by a 55 VDC battery which is recharged by a solar panel
The 111 W solar panel only generates 12 VDC, so an inverter is used to boost the voltage to 55 VDC to
recharge the battery
• The inverter is 90% efficient (output power is 90% of input power)
• After four particularly cloudy days (were the solar panel was unable to recharge the battery), the solar panel is
finally able to recharge the half discharged battery to full in 4.2 hours
What is the rated capacity of the 55 VDC battery (in Ah)? Answer tolerance is 5%.

Expert Solution

Step 1

The 55 V DC battery is half discharged. It takes 4.2 hours to recharge the half-discharged battery. So, it takes 8.4 hours to fully charge. Now, the solar panel output power is 111 W. It is connected to the inverter. The inverter is 90% efficient.

The output of the inverter is,

$\begin{array}{rcl} P_{o u t} & = & η \times P_{i n} \\ = & 90 % \times 111 \\ = & 0.9 \times 111 \\ = & 99.9 W \end{array}$

The inverter output power is used as input to charge the 55 VDC.

The charging current of the battery is,

$\begin{array}{rcl} I & = & \frac{P_{o u t}}{V D C} \\ = & \frac{99.9}{55} \\ = & 1.816 A \end{array}$

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