Chapter 3 Ohm’s Law, Energy, and Power

Chapter 3 Ohm’s Law, Energy, and Power

ElectronicCircuits&Electronics Shao-YuLien 1

3.1 Ohm’s Law

• Ohm determined experimentally that – If the voltage across a resistor is increased, the current through the

resistor will increase– If the voltage across a resistor is decreased, the current through the

resistor will decrease


3.1 Ohm’s Law

• If the voltage is held constant– Less resistance results in more current – More resistance results in less current


3.1 Ohm’s Law

• Ohm’s law states that current is directly proportional to the voltage and inversely proportional to resistance

𝐼 =𝑉𝑅

– I: current in amperes (A)– V: voltage in colts (V)– R: resistance in ohms (Ω)

• Equivalent forms of Ohm’s lawV=IR

𝑅 =𝑉𝐼


3.1 Ohm’s Law

• A graphic aid for Ohm’s law


3.1 Ohm’s Law

Example. Show that if the voltage in the following circuit is increased three times its present value, the current will triple in value.

• With 10 V, the current is

𝐼 =𝑉𝑅=

10𝑉100Ω

= 0.1𝐴


3.1 Ohm’s Law

• With 30 V, the current is

𝐼 =𝑉𝑅=

30𝑉100Ω

= 0.3𝐴


3.2 Applications of Ohm’s Law

Example. Calculate the current in mA for the following circuit

𝐼 = ,-.= /01

2.034= 50×1078A=50 mA



Example. How much voltage will be measured across the resistor in the following circuit?

VR=IR=(5 mA)(56 Ω)=(5×10-3 A)(56 Ω)=280 mA



Example. An automotive lamp draws 2 A from the 13.2 V battery. What is the resistance of the bulb?

𝑅 =𝑉𝐼=13.2V2.0A

= 6.6Ω


3.3 Energy and Power

• Energy is the ability to do work, and power is the rate at which energy is used.

𝑃 =𝑊𝑡

– P: power in watts (W)– W: energy in joules (J)– t: time in seconds (s)

• One watt is the amount of power when one joule of energy is used in one second.



Example. An amount of energy equal to 100 J is used in 5 s. What is the power in W?

𝑃 =𝑊𝑡=100J5s

= 20W



• Kilowatt-hour (kWh) unit of energy: use 1000 W of power for 1 h

Example. Determine the number of kWh for each of the following energy consumption.• 1400 W for 1 h• 2500 W for 2 h• 100,000 W for 5 h

• 1400 W=1.4 kWW=Pt=(1.4 kW)(1 h)=1.4 kWh



• 2500 W=2.5 kWW=Pt=(2.5 kW)(2 h)=5 kWh

• 100,000 W=100 kWW=Pt=(100 kW)(5 h)=500 kWh


3.4 Power in Electric Circuit

• The amount of power dissipated in an electrical circuit is dependent on the amount of resistance and on the amount of current

P=I2R=VI=,C

.



Example. Calculate the power in each of the following three circuits

• (a) The power is determined by P=VI=(10 V)(2 A)=20 W



• (b) The power is determined by P=I2R=(2 A)2(47 Ω)=188 W

• (c) The power is determined by

𝑃 = ,C

.= (/1)C

204= 2.5W


3.5 The Power Rating of Resistors


• Power rating: the maximum amount of power that a resistor can dissipate without being damaged by excessive heat buildup

• Power rating is not related to the resistance• Power rating is determined by the physical composition, size

and shape of the resistor• The larger the surface area of a resistor, the more power it

can dissipate



• The surface area of a cylindrically shaped resistor is equal to the length (l) times the circumference (c)



1/8 W

1/4 W

1/2 W

1 W


Example. Choose an adequate power rating (1/8 W, 1/4 W, 1/2 W, or 1 W) for each of the metal-film resistors in the following



• For the circuit in (a), the actual power is

𝑃 = ,FC

.= (201)C

2G04= 0.833W

A 1 W resistor should be used.

• For the circuit in (b), the actual power is 𝑃 = 𝐼G𝑅=(10 mA)2(1000 Ω)=0.1 W

A 1/8 W (0.125 W) resistor should be used.


3.6 Energy Consumption and VoltageDrop in a Resistance • The decrease in voltage across the resistor due to a loss of

energy is called a voltage drop• Recall that V=W/Q


Wenter/Q Wexit/Q Wenter>Wexit

3.7 Power Supplies and Batteries

• A power supplier is a device that converts alternative current (ac) to a direct current (dc) voltage

• Batteries are also able to supply dc• Power supply efficiency is the ratio of the output power POUT to

the input power PINPUT

Efficiency= POUT/PIN

• The output power is always not larger than the input power, andsuch power dissipation is called power loss

POUT=PIN-PLOSS



Example. A certain power supplier requires 25 W of input power. It can produce an output power of 20 W. What is the efficiency, and what is the power loss?

Efficiency=POUT/PIN=20 W/25 W=0.8PLOSS=PIN-POUT=25 W-20 W=5 W



• The capacity of a battery is measured in ampere-hours (Ah) • Ampere-hour (Ah) rating of batteries determines the length of

time a battery can deliver a certain amount of current at the rated voltage

• A 12 V automobile battery may be rated for 70 Ah at 3.5 A: it can supply an average of 3.5 A for 20 h at the rated voltage



Example. For how many hours can a battery deliver 2 A if it is rated at 70 Ah?

Let x denote the number of hours a battery can deliver 2 A,70 Ah=(2 A)(x h)

x=(70 Ah)/(2 A)=35 h


Chapter 3 Ohm’s Law, Energy, and Power

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