Chapter 25: Electric Circuits

Resistors in Series and Parallel Resistors in series

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Resistors in Series and Parallel

Resistors in parallel

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Resistors in Series and Parallel

Example 1:

Resistors in Series and Parallel

Example: (cont’d)

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A 4 V/3 12/

A 6 V/2 12/

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Resistors in Series and Parallel

Example: (cont’d)

Kirchhoff’s Rules

Introduction

Loop 1

Loop 2

i i

i

i

i1

i2

i2

• Many practical resistor networks cannot be reduced to simple series-parallel combinations (see an example below).• Terminology:

-A junction in a circuit is a point where three or more conductors meet.-A loop is any closed conducting path.

junction

junction

Kirchhoff’s Rules Kirchhoff’s junction rule

• The algebraic sum of the currents into any junction is zero:

junction any at 0I

Kirchhoff’s Rules

Kirchhoff’s loop rule

• The algebraic sum of the potential differences in any loop, including those associated with emfs and those of resistive elements, must equal zero. loopany for 0V

Kirchhoff’s Rules Rules for Kirchhoff’s loop rule

junction any at 0I

loopany for 0V

Kirchhoff’s Rules Rules for Kirchhoff’s loop rule (cont’d)

Kirchhoff’s Rules Solving problems using Kirchhoff’s rules

Kirchhoff’s Rules Example 1

Kirchhoff’s Rules Example 1 (cont’d)

Kirchhoff’s Rules Example 1 (cont’d)

Find all the currents including directions.

Loop 1

Loop 2

i i

i

i

i1

i2

i2

Kirchhoff’s Rules Example 2

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Electrical Measuring Instruments Galvanometer

To be discussed in a laterclass.

Electrical Measuring Instruments Ammeter

Electrical Measuring Instruments Ammeter (cont’d)

Electrical Measuring Instruments Voltmeter

R-C Circuits Charging a capacitor

R-C Circuits Charging a capacitor (cont’d)

R-C Circuits Charging a capacitor (cont’d)

R-C Circuits Charging a capacitor (cont’d)

R-C Circuits Charging a capacitor (cont’d)

R-C Circuits Discharging a capacitor

R-C Circuits Discharging a capacitor (cont’d)

R-C Circuits Discharging a capacitor (cont’d)

Exercises

Problem 1

20 mA

500 A

Rc=25.0

Rs

The resistance of a galvanometer coil is 25.0 ,and the current required for full-scale deflectionis 500 A.a) Show in a diagram how to convert the galvano-

meter to an ammeter reading 20.0 mA full scale,and compute the shunt resistance.

b) Show how to convert the galvanometer to a voltmeter reading 500 mV full scale, and compute the series resistance.

500 A

Vab=500 mV

Rc=25.0

Rs

a b

a) ammeter

b) voltmeter

a) For a 20-mA ammeter, the two resistance are in parallel: Vc=Vs->IcRc=IsRs->(500 x 10-6 A)(25.0 ) = (20 x 10-3 A – 500 x 10-6 A)Rs-> Rs=0.641 .b) For a 500-mV voltmeter, the resistances are in series: Vab=I(Rc+Rs)->Rs=Vab/I – Rc -> Rs=500 x 10-3 V / 500 x 10-6 A – 25.0 = 975 .

Solution

Exercises Problem 2 20.0 V 5.00

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Exercises Problem 3

Sa b

6.00 F

3.00 F

6.00

3.

00

V=18.0 Va) What is the potential of point a with respectto point b when the switch S is open?

b) Which point, a or b, is at higher potential?Now the switch S is closed.b) What is the final potential of point b?c) How much charge flows through switch S

when it is closed?

a) With an open switch:

Also, there is a current in the left branch:

So

b) Point b is at the higher potential.c) If the switch is closed:d) New charges are:

C. 103.60V) F)(18.0 1000.2( -56 VCQ eq

A. 2.00) 3.00 V)/(6.00 0.18( I

V. -6.00) A)(6.0 (2.0-F) 10C)/(6.0 106.3(

/6-5

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V. 6.00) A)(3.00 00.2( ab VV

C. 103.60C) 10(-7.20-C 1060.3

C. 101.80C) 10(1.80-C 1060.3

C. 10-7.20V) F)(-12.0 1000.6(

C. 101.80V) F)(6.0 1000.3(

5-5-56

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Q

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CVQThe total charge flowingthrough the switch is5.40 x 10-5 C.

Solution