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Slides on Measuring Voltage and Current | ES A309, Study notes of Engineering

University of Alaska AnchorageEngineering
Prof. Jeffrey A. Miller

Material Type: Notes; Professor: Miller; Class: Elements of Electrical Engineering; Subject: Engineering Science ; University: University of Alaska - Anchorage; Term: Unknown 1989;

Typology: Study notes

2009/2010

Uploaded on 03/28/2010

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ES309
Elements of Electrical Engineering
Lecture #4
Jeffrey Miller, Ph.D.
Outline
• Chapter 3.5-3.7
Measuring Voltage and Current
• Ammeter
– Instrument designed to measure current
– Placed in series with the circuit element whose
current is being measured
– Resistance should be 0Ω
• Voltmeter
– Instrument designed to measure voltage
– Placed in parallel with the element whose
voltage is being measured
– Resistance should be ∞Ω
Sample Resistor Problem
Find the voltage and current across each resistor
pf3
pf4
pf5

Partial preview of the text

Download Slides on Measuring Voltage and Current | ES A309 and more Study notes Engineering in PDF only on Docsity!

ES

Elements of Electrical Engineering

Lecture

Jeffrey Miller, Ph.D.

Outline

• Chapter 3.5-3.

Measuring Voltage and Current

• Ammeter

– Instrument designed to measure current

– Placed in series with the circuit element whose

current is being measured

– Resistance should be 0Ω

• Voltmeter

– Instrument designed to measure voltage

– Placed in parallel with the element whose

voltage is being measured

– Resistance should be ∞Ω

Sample Resistor Problem

Find the voltage and current across each resistor

Find equivalent circuit

Req = 7kΩ + 5kΩ

= 12kΩ

Find equivalent circuit

Req = 3kΩ + 8kΩ

= 11kΩ

Find equivalent circuit

1 / Req = 1 / 6kΩ + 1 / 12kΩ = 6kΩ + 12kΩ / (6kΩ * 12kΩ) = 18kΩ / (72kΩ)^2 = 1 / 4kΩ Req = 4kΩ

Find equivalent circuit

Req = 11kΩ + 4kΩ

= 15kΩ

Current over 12kΩ resistor

1 / Req = 1 / 6kΩ + 1 / 12kΩ = 12kΩ + 6kΩ / (6kΩ * 12kΩ) = 18kΩ / 72kΩ = 1/4kΩ Req = 4kΩ i12kΩ = (Req / R12kΩ)* i = (4kΩ / 12kΩ) * 4/5mA = 4/15 mA

Current over 3kΩ and 8kΩ resistors

The current over two resistors in series is the same, so

the current on the 3kΩ and 8kΩ resistors is the same as

the current over the 11kΩ resistor, which is 4/5mA

Current over 5kΩ and 7kΩ resistors

The current over two resistors in series is the same, so

the current on the 5kΩ and 7kΩ resistors is the same as

the current over the 12kΩ resistor, which is 4/15mA

Current across all resistors

Voltage across all resistors

v = iR v10kΩ = 6/5mA * 10kΩ = 12V v3kΩ = 4/5mA * 3kΩ = 2.4V v8kΩ = 4/5mA * 8kΩ = 6.4V v6kΩ = 8/15mA * 6kΩ = 3.2V v5kΩ = 4/15mA * 5kΩ = 1.3V v7kΩ = 4/15mA * 7kΩ = 1.87V

Wheatstone Bridge Example

Assume there is no current, and therefore no

voltage, in the detector (aka galvonometer)

Find the value of Rx

Wheatstone Bridge Example

i 1 = i 3 i 2 = ix

Since there is no current or voltage drop over the detector, points a and b are at the same potential, giving:

i 3 R 3 = ixRx i 1 R 1 = i 2 R 2

Then, combining the previous equations give:

i 1 R 3 = i 2 Rx

Wheatstone Bridge Example

i 1 R 3 = i 2 Rx

Dividing this expression by i 1 R 1 = i 2 R 2

gives:

R 3 /R 1 = Rx/R 2

Rx = R 2 R 3 / R 1