# 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. ### Principles of Instrumental Analysis

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

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

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

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

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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