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Exercises on electrical circuits | PHYS 142, Quizzes of Physics

University of Illinois - ChicagoPhysics

Material Type: Quiz; Class: General Physics II (Electricity and Magnetism); Subject: Physics; University: University of Illinois - Chicago; Term: Summer 2009;

Typology: Quizzes

Pre 2010

Uploaded on 08/16/2009

koofers-user-9z8
koofers-user-9z8 🇺🇸

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QUIZ#3 (PHYS-142)
July 13, 2009 (Monday) Time: 12:00 pm – 1:40 pm
Name: Roll No:
Q.1
a) An ideal voltmeter has infinite resistance, so there would be NO current through the
resistor.0.2
b) 
;V0.5
ε
ab
V
since there is no current there is no voltage lost over the
internal resistance.
c) The voltmeter reading is therefore 5.0 V since with no current flowing, it measures
the terminal voltage of the battery.
Q.2 The circuit is shown below. Find (a) the current through
the circuit (magnitude & direction) (b) the terminal
voltage Vab (c) Potential difference Vac .
a) The current is counterclockwise, because the 16 V battery determines the direction of
current flow. Its magnitude is given by:
A.47.0
0.94.10.56.1
V0.8V0.16
R
I
ε
b) 
.V2.15)A47.0)(6.1(V0.16
ab
V
c) 
V.0.11V0.8)A47.0)(4.1()A47.0)(0.5(
ac
V
pf3
pf4

Partial preview of the text

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QUIZ#3 (PHYS-142)

July 13, 2009 (Monday) Time: 12:00 pm – 1:40 pm

Name: Roll No:

Q.

a) An ideal voltmeter has infinite resistance, so there would be NO current through the

  1. 0  resistor.

b) Vab^  ε ^5.^0 V;since there is no current there is no voltage lost over the

internal resistance.

c) The voltmeter reading is therefore 5.0 V since with no current flowing, it measures

the terminal voltage of the battery.

Q.2 The circuit is shown below. Find (a) the current through

the circuit (magnitude & direction) (b) the terminal

voltage Vab (c) Potential difference Vac.

a) The current is counterclockwise, because the 16 V battery determines the direction of

current flow. Its magnitude is given by:

0. 47 A.

16. 0 V 8. 0 V

R

I

b) Vab ^16.^0 V (^1.^6 )(^0.^47 A)^15.^2 V.

c) Vac (^5.^0 )(^0.^47 A)(^1.^4 )(^0.^47 A)^8.^0 V^11.^0 V.

Q.3 A plastic tube 25 m long and 4 cm in diameter is dipped

into a silver solution, depositing a layer of silver 0.1 mm thick

uniformly over the outer surface of the tube. If this coated

tube is then connected across a 12 V battery, what will be the

current? Assume that the resistivity of silver is 1.47 x 10-8^ Ω.m.

Sol: r = 2.00 cm

T = 0.100 mm

ρll V πrTrT ρll VA ρll A V R V I ( 2 )     (1.47 10 m)(25.0 m) ( 12 V)( 2 )( 2. 00 10 m)(0.100 10 m) 8 2 3   

πrT ^   410 A

Q.4 Find (a) the current through the battery and through each

resistor in the circuit shown below. (b) the equivalent

resistance connected to the battery.

Sol: a) Using the currents as defined on the circuit diagram below we obtain three

equations to solve for the currents:

Q6. In the circuit shown in the figure C=5.0 μF,F,

Charge on the capacitor a long time after the switch is moved

to position 2? (b) after the switch has been in position 2 for 3

ms, the charge on the capacitor is measured to be 110 μF,C.

What is the value of the resistance? (c) How long after the

switch is moved to position 2 will the charge on the capacitor

be equal to 99% of the final value found in part (a)?

Sol: a) Q  CV ( 5. 90  10 ^6 F)( 28. 0 V) 1. 65  10 ^4 C.

b) (^1 )^1 ln( 1 / ).

/ / C q Q t R Q q q Q e t RC e t RC           

After 463.

( 5. 90 10 F)(ln( 1 110 / 165 )) 3 10 s 3 10 s : 6 3 (^3)    

t^ ^ R

c) If the charge is to be 99% of final value:

( 463 )( 5. 90 10 F)ln( 0. 01 ) 0. 0126 s. ( 1 ) ln( 1 / ) 6 /           e^ ^ t RC q Q Q q (^) t RC