This

**preview**shows half of the first page. to view the full**3 pages of the document.**Solutions to Homework 2

3.3 Values given are: V = 20 Volts,

== 1000

21 RR

,

=9500

m

R

.

(i) Let the current flowing round the circuit with the voltmeter in position be I.

The resistance

AB

R

is composed of two resistances

2

R

and

m

R

in parallel. The value of

AB

R

can be calculated

from the formula for the effective resistance of two resistors in parallel:

mAB RRR

111

2

+=

i.e.

m

m

AB RR

RR

R+

=

2

2

Substituting in for the given numerical values of

2

R

and

m

R

:

=

+

=76.904

95001000

95001000

AB

R

According to Ohm’s law, I is given by:

AB

RR

V

I+

=

1

Then the voltage across AB can be calculated from Ohm’s law as:

5.9

76.9041000

76.90420

1

=

+

=

+

==

AB

AB

ABAB RR

RV

RIV

volts

Thus the reading on the voltmeter is 9.5 volts.

(ii) Let the current flowing tround the circuit without the voltmeter in position be

I

.

I

can be calculated simply from Ohm’s law as:

21 RR

V

I+

=

The voltage across

2

V

can be calculated as

21

2

2RR

RV

RIV +

==

Substituting in numerical values:

0.10

10001000

100020 =

+

=

V

volts.

Thus, the voltage across AB when the voltmeter is not loading the circuit is 10.0 volts.

(iii) The measurement error due to the loading effect of the voltmeter is simply

AB

VV −

, i.e. 0.5 volts.

3.26 Part (i)

Given data: = mean temperature = 75C ; = standard deviation = 2.15

Applying

( )

−= xz

; For

3256.2,70 −== zx

( ) ( )

3256.213256.23256.23256.2170170 +−=−=−=−−=−= FFzPzPxPxP

Using error function (z-function or gaussian) table:

( )

1.099.013256.21 =−=− F

Thus, the temperature is less than 70C for 1% of the time.

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