BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213
BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

Solutions

Chapter 2, Problem 2.1QAP
Interpretation Introduction

Interpretation:

A suitable combination of the resistors with the indicated voltage needs to be determined.

Concept Introduction :

The resistors in series from a voltage divider in that fraction of the total voltage appear across each resistor.

Expert Solution

Answer to Problem 2.1QAP

A suitable combination of the resistors that would give the indicated voltages are R1=500Ω,R2=2kΩ,R3=2.5kΩ. .

Explanation of Solution

Consider the diagram, all the resistors ( R1,R2,and R3 ) are in series.

The resistors in series from a voltage divider in that fraction of the total voltage appear across each resistor.

Applying Ohm’s law,from A to B, the voltage across the resistor R1 is:

  V1=I1R1

The fraction of the total voltage appears across resistor R1 is,

  V1V=IR1I( R 1 + R 2 + R 3 )=R1( R 1 + R 2 + R 3 )=R1Rs

Here

  Rs=R1+R2+R3

Given

  V=10.0V,V1=1.0V,V2=4.0V

Therefore,

  V1V=1.0V10.0V=110=R1RsRs=10R1

Similarly,

  V2V=R2Rs4.0V10.0V=R2RsRs=104R2

Substitute the value Rs=10R1

Therefore,

  10R1=104R2R1R2=14

Here, R1=500Ω,R2=2kΩ=2000Ω .

Hence, it is shown that

:

  R1R2=5002000=14

In the case of series resistors, the value of Rs is:

  Rs=R1+R2+R3

  Rs=500+2000+R3V1Vs=R1Rs

Or,

  110=500500+2000+R3R3=2500Ω=2.5kΩ

Interpretation Introduction

Interpretation:

The IR drop across R3 needs to be determined.

Concept Introduction :

According to the Kirchhoff’s law, the algebraic sum of the voltage around the closed path is zero.

Expert Solution

Answer to Problem 2.1QAP

IR drop across R3 is 5.0V.

Explanation of Solution

According to the Kirchhoff’s law, the algebraic sum of the voltage around the closed path is zero.

It is represented as

  V=V1+V2+V3

Given,

  V=10.0V,V1=1.0V,V2=4.0V

  10.0V=1.0V+4.0V+V3V3=5.0

IR drop across R3 is 5.0V.

Interpretation Introduction

Interpretation:

The current drawn from the source needs to be determined.

Concept Introduction : According to the Ohms law V=IR .

Here, V is voltage, I is current, and R is resistance.

Expert Solution

Answer to Problem 2.1QAP

Current drawn from the source is 0.002A.

Explanation of Solution

According to Ohm’s law,

  V=IR

Consider the series resistors,

  R=R1+R2+R3=500Ω+2000Ω+2500Ω=5000ΩI=VR=10V5000Ω=0.002A

Thus, the current drawn from the source is 0.002A.

Interpretation Introduction

Interpretation:

The dissipated power needs to be calculated.

Concept Introduction:

The power dissipated in the circuit is the product of current and potential difference across the element.

Expert Solution

Answer to Problem 2.1QAP

The power dissipated by the circuit is 0.02W.

Explanation of Solution

Consider the power is:

  P=VIP=0.002 A×10.0 V=0.02 W

Hence, the dissipated power is 0.02W.

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