COLLEGE PHYSICS (OER)
COLLEGE PHYSICS (OER)
1st Edition
ISBN: 9781947172012
Author: DIRKS
Publisher: OpenStax
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Chapter 21, Problem 50PE
To determine

(a)

The current for the given circuit

Expert Solution
Check Mark

Answer to Problem 50PE

  I=1.5848mA

Explanation of Solution

Given information:

The voltage of alkaline cell is 1.585 V.

An internal resistance is 0.100Ω.

Voltmeter is used with 1kΩ

The given figure is

COLLEGE PHYSICS (OER), Chapter 21, Problem 50PE , additional homework tip  1

Introduction:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

Ohms law states that at constant temperature potential difference V across the terminal is directly proportional to current I flowing through the circuit

  V=IR

Where R is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  Req=R1+R2+............+RN

It is known that

Ohms law states that at constant temperature potential difference V across the terminal is directly proportional to current V flowing through the circuit

  V=IR

Where R is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  Req=R1+R2+............+RN

It is given that

  V=1.585 V

Resistors are connected in series so

  Req=0.100+1000(1kΩ=1000Ω)=1000.1

Substitute all the values in V=IR

  I=VR eqI=1.5851000.1I=0.0015848I=1.5848mA

Conclusion:

The current for the given circuit is I=1.5848mA.

To determine

(b)

The terminal voltage of the given circuit

Expert Solution
Check Mark

Answer to Problem 50PE

  1.5848V

Explanation of Solution

Given information:

The voltage of alkaline cell is 1.585 V.

An internal resistance is 0.100Ω.

Voltmeter is used with 1kΩ

The given figure is

COLLEGE PHYSICS (OER), Chapter 21, Problem 50PE , additional homework tip  2

Introduction:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

Ohms law states that at constant temperature potential difference V across the terminal is directly proportional to current V flowing through the circuit

  V=IR

Where R is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  Req=R1+R2+............+RN

It is known that

Ohms law states that at constant temperature potential difference V across the terminal is directly proportional to current V flowing through the circuit

  V=IR

Where R is a constant value called as resistance

It is given that

  I=1.5848mA

  R=

  1kΩ

Substitute all the values in V=IR

  V=IRV=1.5848×1=1.5848V

Conclusion:

The terminal voltage of the given circuit is 1.5848V.

To determine

(c)

The ratio of the measured terminal voltage to the emf for the given circuit

Expert Solution
Check Mark

Answer to Problem 50PE

  0.9998

Explanation of Solution

Given information:

The voltage of alkaline cell is 1.585 V.

An internal resistance is 0.100Ω.

Voltmeter is used with 1kΩ

The given figure is

COLLEGE PHYSICS (OER), Chapter 21, Problem 50PE , additional homework tip  3

Introduction:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

Ohms law states that at constant temperature potential difference V across the terminal is directly proportional to current V flowing through the circuit

  V=IR

Where R is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  Req=R1+R2+............+RN

The terminal voltage of the given circuit is 1.5848V.

The given emf is 1.585 V.

The ratio of the measured terminal voltage to the emf for the given circuit is

  measured terminal voltagegiven emf=1.58481.585=0.9998

Conclusion:

The ratio of the measured terminal voltage to the emf for the given circuit is 0.9998.

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Chapter 21 Solutions

COLLEGE PHYSICS (OER)

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(a) What is the unknown emfx in a potentiometer...Ch. 21 - Suppose you want to measure resistances in the...Ch. 21 - The timing device in an automobile’s intermittent...Ch. 21 - A heart pacemaker fires 72 times a minute, each...Ch. 21 - The duration of a photographic flash is related to...Ch. 21 - A 2.00- and a 7.50-F capacitor can be connected in...Ch. 21 - After two time constants, what percentage of the...Ch. 21 - A 500- resistor, an uncharged 1.50-F capacitor and...Ch. 21 - A heart defibrillator being used on a patient has...Ch. 21 - An ECG monitor must have an RC time constant less...Ch. 21 - Prob. 71PECh. 21 - Using the exact exponential treatment, find how...Ch. 21 - Using the exact exponential treatment, find how...Ch. 21 - Integrated Concepts If you wish to take a picture...Ch. 21 - Integrated Concepts A flashing lamp in a Christmas...Ch. 21 - Integrated Concepts A 160F capacitor charged to...Ch. 21 - Unreasonable Results (a) Calculate the capacitance...Ch. 21 - Construct Your Own Problem Consider a camera's...Ch. 21 - 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