Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
Question
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Chapter 2, Problem 2.6QAP
Interpretation Introduction

(a)

Interpretation:

The power dissipated between the points 1 and 2 is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The power in dissipated in the electrical circuit is defined by I2R, where the term I is the current flowing through the circuit and the R is the equivalent resistance of the electrical circuit across the source.

Expert Solution
Check Mark

Answer to Problem 2.6QAP

The power dissipated between the points 1 and 2 is 0.0514W.

Explanation of Solution

The below figure represents the electrical circuit.

Principles of Instrumental Analysis, Chapter 2, Problem 2.6QAP

Write the expression for the equivalent resistance between the point 1 and 2.

R12=RBRCRB+RC ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

R23=RDRERD+RE ....... (II)

Write the expression for the total equivalent resistance.

R=RA+R12+R23 ....... (III)

Write the expression for the current flowing through the circuit.

I=VR ....... (IV)

Here, V is the voltage of the voltage source.

Write the expression for the power dissipated between the points 1 and 2.

P12=I2R12 ....... (V)

Substitute 3.0 for RB and 4.0 for RC in the Equation (I).

R12=(3.0)(4.0)(3.0)+(4.0)R12=1.714

Substitute 2.0 for RD and 1.0 for RE in the Equation (II).

R23=(2.0)(1.0)(2.0)+(1.0)R23=0.667

Substitute 1.714 for R12, 0.667 for R23 and 2.0 for RA in the Equation (III).

R=(2.0)+(1.714)+(0.667)R=4.381

Substitute 4.381 for R and 24V for the V in the Equation (IV).

I=24V(4.381)(103Ω1)I=5.478×103A

Substitute 5.478×103A for I and 1.714 for R12 in the Equation (V).

P12=(5.478×103A)2(1.714)(103Ω1)P12=0.0514W

Interpretation Introduction

(b)

Interpretation:

The current drawn from the source is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is to evaluating the equivalent resistance, current through the circuit and power consumed by the circuit.

The current drown the from the source is define as I=VR, where V represents the voltage of the source and R represents the equivalent resistance of the circuit.

Expert Solution
Check Mark

Answer to Problem 2.6QAP

The current drown from the source is 5.478mA.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

R12=RBRCRB+RC ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

R23=RDRERD+RE ....... (II)

Write the expression for the total equivalent resistance.

R=RA+R12+R23 ....... (III)

Write the expression for the current flowing through the circuit.

I=VR ....... (IV)

Here, V is the voltage of the voltage source.

Substitute 3.0 for RB and 4.0 for RC in the Equation (I).

R12=(3.0)(4.0)(3.0)+(4.0)R12=1.714

Substitute 2.0 for RD and 1.0 for RE in the Equation (II).

R23=(2.0)(1.0)(2.0)+(1.0)R23=0.667

Substitute 1.714 for R12, 0.667 for R23 and 2.0 for RA in the Equation (III).

R=(2.0)+(1.714)+(0.667)R=4.381

Substitute 4.381 for R and 24V for the V in the Equation (IV).

I=24V(4.381)(103Ω1)I=(5.478×103A)(1mA103A)I=5.478mA

Interpretation Introduction

(c)

Interpretation:

The voltage drop across the resistor RA is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The voltage drop across the resistor is defined as V=IR, where I is the current through the resistor and the R is the resistance of the resistor.

Expert Solution
Check Mark

Answer to Problem 2.6QAP

The voltage drop across the resistance RA is 10.956V.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

R12=RBRCRB+RC ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

R23=RDRERD+RE ....... (II)

Write the expression for the total equivalent resistance.

R=RA+R12+R23 ....... (III)

Write the expression for the current flowing through the circuit.

I=VR ....... (IV)

Here, V is the voltage of the voltage source.

Write the expression for the voltage drop across the resistor RA.

VA=IRA ....... (V)

Substitute 3.0 for RB and 4.0 for RC in the Equation (I).

R12=(3.0)(4.0)(3.0)+(4.0)R12=1.714

Substitute 2.0 for RD and 1.0 for RE in the Equation (II).

R23=(2.0)(1.0)(2.0)+(1.0)R23=0.667

Substitute 1.714 for R12, 0.667 for R23 and 2.0 for RA in the Equation (III).

R=(2.0)+(1.714)+(0.667)R=4.381

Substitute 4.381 for R and 24V for the V in the Equation (IV).

I=24V(4.381)(103Ω1)I=5.478×103A

Substitute 5.478×103A for I and 2.0 for RA in the Equation (V).

VA=(5.478×103A)(2.0)(103Ω1)VA=10.956V

Interpretation Introduction

(d)

Interpretation:

The voltage drop across the resistance RD is to be stated.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The voltage drop across the resistor is defined as V=IR, where I is the current through the resistor and the R is the resistance of the resistor.

Expert Solution
Check Mark

Answer to Problem 2.6QAP

The voltage drop across the resistor RD is 3.707V.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

R12=RBRCRB+RC ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

R23=RDRERD+RE ....... (II)

Write the expression for the total equivalent resistance.

R=RA+R12+R23 ....... (III)

The voltage drop across the resistor RD is the voltage drop across the resistor R23.

Write the expression for the voltage drop across the resistor RD.

VD=(R23R)V ....... (IV)

Substitute 3.0 for RB and 4.0 for RC in the Equation (I).

R12=(3.0)(4.0)(3.0)+(4.0)R12=1.714

Substitute 2.0 for RD and 1.0 for RE in the Equation (II).

R23=(2.0)(1.0)(2.0)+(1.0)R23=0.667

Substitute 1.714 for R12, 0.667 for R23 and 2.0 for RA in the Equation (III).

R=(2.0)+(1.714)+(0.667)R=4.381

Substitute 0.667 for R23, 4.381 for R and 24V for V in the Equation (IV).

VD=(0.6674.318)(24V)VD=3.707V

Interpretation Introduction

(e)

Interpretation:

The potential difference between the points 5 and 4.

Concept introduction:

The resistive electrical circuit is the circuit which contains the voltage source and the resistor. The analysis of the circuit is done to evaluate the equivalent resistance, current through the circuit and power consumed by the circuit.

The voltage drop across the resistor is defined as V=IR, where I is the current through the resistor and the R is the resistance of the resistor.

Expert Solution
Check Mark

Answer to Problem 2.6QAP

The potential difference across the points 5 and 4 is 13.044V.

Explanation of Solution

Write the expression for the equivalent resistance between the point 1 and 2.

R12=RBRCRB+RC ....... (I)

Write the expression for the equivalent resistance between the point 2 and 3.

R23=RDRERD+RE ....... (II)

Write the expression for the total equivalent resistance.

R=RA+R12+R23 ....... (III)

Write the expression for the current flowing through the circuit.

I=VR ....... (IV)

Here, V is the voltage of the voltage source.

Write the expression for the voltage drop across the resistor RA.

VA=IRA ....... (V)

Write the expression for the potential difference across the points 5 and 4.

V54=V12+V23 ....... (VI)

Write the expression for the total voltage.

V=VA+V12+V23 ....... (VII)

Substitute 3.0 for RB and 4.0 for RC in the Equation (I).

R12=(3.0)(4.0)(3.0)+(4.0)R12=1.714

Substitute 2.0 for RD and 1.0 for RE in the Equation (II).

R23=(2.0)(1.0)(2.0)+(1.0)R23=0.667

Substitute 1.714 for R12, 0.667 for R23 and 2.0 for RA in the Equation (III).

R=(2.0)+(1.714)+(0.667)R=4.381

Substitute 4.381 for R and 24V for the V in the Equation (IV).

I=24V(4.381)(103Ω1)I=5.478×103A

Substitute 5.478×103A for I and 2.0 for RA in the Equation (V).

VA=(5.478×103A)(2.0)(103Ω1)VA=10.956V

Substitute V12+V23 for V54 in the Equation (VII).

V=VA+V54V54=VVA

Substitute 10.956V for VA and 24V for V.

V54=(24V)(10.956V)V54=13.044V

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