Glencoe Physics: Principles and Problems, Student Edition
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN: 9780078807213
Author: Paul W. Zitzewitz
Publisher: Glencoe/McGraw-Hill
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Chapter 23.1, Problem 12PP

a.

To determine

Theequivalent resistance of the circuit.

a.

Expert Solution
Check Mark

Answer to Problem 12PP

The equivalent resistance in the circuit is 11.9 kΩ.

Explanation of Solution

Given:

Three resistors of value 3.3 kΩ, 4.7 kΩ and 3.9 kΩ are connected in series across a 12-V battery.

Formula used:

For a set of resistors R1 , R2 , R3 in series, the equivalent resistance can be calculated as,

  Req=R1+R2+R3+... …… (1)

Calculation Consider the circuit shown in Figure 1. Here, the generator is supplying a voltage of 12-V and is loaded with three resistances in series, 3.3 kΩ, 4.7 kΩ and 3.9 kΩ

  Glencoe Physics: Principles and Problems, Student Edition, Chapter 23.1, Problem 12PP

Figure 1

The equivalent resistance can be calculated using (1) as,

  Req=3.3+4.7+3.9=11.9 kΩ

Conclusion:

The equivalent resistance of the circuit is 11.9 kΩ .

b.

To determine

The current in the circuit.

b.

Expert Solution
Check Mark

Answer to Problem 12PP

The current through the circuit is 1.01 mA.

Explanation of Solution

Given:

Three resistors of value 3.3 kΩ, 4.7 kΩ and 3.9 kΩ are connected in series across a 12-V battery.

Formula used:

For the resistance of the circuit and supply voltage, the current can be calculated using the Ohm’s law as,

  V=IRI=VR …… (2)

Calculation:

The current flowing through the circuit can be calculated by using the relation in (2), by substituting the total resistance obtained in the part (a) as

  I=1211.9×103=1.01 mA

Conclusion:

The current through the circuit is 1.01 mA.

c.

To determine

Thetotal potential difference across the three resistors.

c.

Expert Solution
Check Mark

Answer to Problem 12PP

The total potential difference across the three resistors is 12 V

The potential differences across the three resistors are

  V3.3k=3.333 V , V4.7k=4.747 V

  V3.9k=3.939 V

Explanation of Solution

Given:

Three resistors of value 3.3 kΩ, 4.7 kΩ and 3.9 kΩ are connected in series across a 12-V battery.

Formula used:

The total potential difference across the three resistors is the supply voltage itself. It can also be expressed as the product of the current through the circuit and the total resistance given by,

  V=IR

Calculation The total potential difference across the resistors can be expressed using (1) as,

  V=11.9 ×103×1.01×103=12 V

The individual potential difference across the resistors can be calculated using (1) as follows.

  V3.3k=3.3 ×103×1.01×103=3.333 V

  V4.7k=4.7 ×103×1.01×103=4.747 V

  V3.9k=3.9 ×103×1.01×103=3.939 V

Conclusion:

Therefore, the total potential difference across the three resistors is 12 V .

The potential differences across the three resistors are

  V3.3k=3.333 V , V4.7k=4.747 V

  V3.9k=3.939 V

Chapter 23 Solutions

Glencoe Physics: Principles and Problems, Student Edition

Ch. 23.1 - Prob. 11PPCh. 23.1 - Prob. 12PPCh. 23.1 - Prob. 13PPCh. 23.1 - Prob. 14PPCh. 23.1 - Prob. 15PPCh. 23.1 - Prob. 16PPCh. 23.1 - Prob. 17PPCh. 23.1 - Prob. 18SSCCh. 23.1 - Prob. 19SSCCh. 23.1 - Prob. 20SSCCh. 23.1 - Prob. 21SSCCh. 23.1 - Prob. 22SSCCh. 23.1 - Prob. 23SSCCh. 23.1 - Prob. 24SSCCh. 23.2 - Prob. 25PPCh. 23.2 - Prob. 26PPCh. 23.2 - Prob. 27PPCh. 23.2 - Prob. 28SSCCh. 23.2 - Prob. 29SSCCh. 23.2 - Prob. 30SSCCh. 23.2 - Prob. 31SSCCh. 23.2 - Prob. 32SSCCh. 23.2 - Prob. 33SSCCh. 23.2 - Prob. 34SSCCh. 23.2 - Prob. 35SSCCh. 23 - Prob. 36ACh. 23 - Prob. 37ACh. 23 - Prob. 38ACh. 23 - Prob. 39ACh. 23 - Prob. 40ACh. 23 - Prob. 41ACh. 23 - Prob. 42ACh. 23 - Prob. 43ACh. 23 - Prob. 44ACh. 23 - Prob. 45ACh. 23 - Prob. 46ACh. 23 - Prob. 47ACh. 23 - Prob. 48ACh. 23 - Prob. 49ACh. 23 - Prob. 50ACh. 23 - Prob. 51ACh. 23 - Prob. 52ACh. 23 - Prob. 53ACh. 23 - Prob. 54ACh. 23 - Prob. 55ACh. 23 - Prob. 56ACh. 23 - Prob. 57ACh. 23 - Prob. 58ACh. 23 - Prob. 59ACh. 23 - Prob. 60ACh. 23 - Prob. 61ACh. 23 - Prob. 62ACh. 23 - Prob. 63ACh. 23 - Prob. 64ACh. 23 - Prob. 65ACh. 23 - Prob. 66ACh. 23 - Prob. 67ACh. 23 - Prob. 68ACh. 23 - Prob. 69ACh. 23 - Prob. 70ACh. 23 - Prob. 71ACh. 23 - Prob. 72ACh. 23 - Prob. 73ACh. 23 - Prob. 74ACh. 23 - Prob. 75ACh. 23 - Prob. 76ACh. 23 - Prob. 77ACh. 23 - Prob. 78ACh. 23 - Prob. 79ACh. 23 - Prob. 80ACh. 23 - Prob. 81ACh. 23 - Prob. 82ACh. 23 - Prob. 83ACh. 23 - Prob. 84ACh. 23 - Prob. 85ACh. 23 - Prob. 86ACh. 23 - Prob. 87ACh. 23 - Prob. 88ACh. 23 - Prob. 89ACh. 23 - Prob. 90ACh. 23 - Prob. 91ACh. 23 - Prob. 92ACh. 23 - Prob. 93ACh. 23 - Prob. 94ACh. 23 - Prob. 95ACh. 23 - Prob. 96ACh. 23 - Prob. 97ACh. 23 - Prob. 98ACh. 23 - Prob. 99ACh. 23 - Prob. 101ACh. 23 - Prob. 102ACh. 23 - Prob. 103ACh. 23 - Prob. 104ACh. 23 - Prob. 105ACh. 23 - Prob. 1STPCh. 23 - Prob. 2STPCh. 23 - Prob. 3STPCh. 23 - Prob. 4STPCh. 23 - Prob. 5STPCh. 23 - Prob. 6STPCh. 23 - Prob. 7STPCh. 23 - Prob. 8STPCh. 23 - Prob. 9STPCh. 23 - Prob. 10STP

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