circuits - the burns home page practice problems ppt.pdfthree resistors are connected to a 10-v ......

24/08/2010

1

Circuits

Practice Questions

Review Formula’s

BB

B

LR

A

V IR

22 V

P IV I RR

Ir R

E Pt

RC

1

1

1 1 1...

...

P n

s n

R R R

R R R

0 1t

RCQ t Q e

221 1 1

2 2 2

QU QV CV

C

0

t

RCI t I e

1

1

1 1 1...

...

S n

P n

C C C

C C C

Q CV

0

t

RCQ t Q e

Charging

Discharging

A wire made of brass and another wire made of silver have the same

length, but the diameter of the brass wire is 4 times the diameter of the

silver wire. The resistivity of brass is 5 times greater than the resistivity of .

Silver. If RB denotes the resistance of the brass wire and RS denotes the

resistance of the silver wire, which of the following is true?

Question

a) RB=5/16 RS

b) RB=4/5 RS

c) RB=5/4 RS

d) RB=5/2 RS

e) RB=16/5 RS

BB

B

LR

A

2

5

4

5

16

s

s

L

A

Let ρs denote the resistivity of silver and let As

denote the cross-sectional area of the silver wire.

For a ohmic conductor, doubling the voltage without changing the

resistance will cause the current to?

Question

a) Decrease by a factor of 4

b) Decrease by a factor of 2

c) Remain unchanged

d) Increase by a factor of 2

e) Increase by a factor of 4

VI

R

Therefore doubling

the voltage, doubles

the current.

24/08/2010

2

If a 60 watt light bulb operates at a voltage of 120V, what is the resistance

of the bulb?

Question

a) 2Ω

b) 30Ω

c) 240Ω

d) 720Ω

e) 7200Ω

2VP

R

22 120240

60

VVR

P W

A battery whose emf is 40V has an internal resistance of 5Ω. If this battery

is connect to a 15Ω resistor R, what will the voltage drop across R be?

Question

a) 10V

b) 30V

c) 40V

d) 50V

e) 70V

Ir R

402

5 15

VI A

V IR

2 15 30V IR A V

Question

Three resistors are connected to a 10-V battery as shown below. What is

the current through the 2.0 Ω resistor?

a) 0.25A

b) 0.50A

c) 1.0A

d) 2.0A

e) 4.0A

4.0Ω

4.0Ω

2.0Ωε=10V

10

10

1

V IR

VI

R

V

A

1 ...s nR R R

V IR

Since all resistors are in series,

the amount of current that

passes through any one of them

is the same. So we need to

simply the circuit to determine

that current.

1 ...

4 4 2

10

s nR R R

Determine the equivalent resistance between points a and b?

Question

a) 0.167Ω

b) 0.25 Ω

c) 0.333 Ω

d) 1.5 Ω

e) 2 Ω

1

13

1 1

12 4

R

1 2

1 1 1

PR R R

12Ω

4Ω

3Ω

3Ω

2 3 3 6R

3

1

1 1

6 3

2R

24/08/2010

3

Three identical light bulbs are connected to a source of emf, as shown in

the diagram above. What will happen if the middle bulb burns out?

Question

a) All the bulbs will go out

b) The light intensity of the other two bulbs will decrease (but they won’t go

out).

c) The light intensity of the other two bulbs will increase.

d) The light intensity of the other two bulbs will remain the same.

e) More current will be drawn from the source emf.

If each bulb has a resistance of

R, then each individual bulb will

draw ε/R. This will be

unchanged if any individual bulb

goes out. (less current will be

drawn from the battery, but the

same amount of current will

pass through each bulb)

Question

An ideal battery is hooked to a light bulb with wires. A

second identical light bulb is connected in parallel to the

first light bulb. After the second light bulb is connected, the

current from the battery compared to when only one bulb

was connected.

a) Is Higher

b) Is Lower

c) Is The Same

d) Don’t know

Bulbs in parallel are like resistors in parallel.

Therefore since the total resistance of parallel

resistors is lower, and the voltage ins the same,

then the current must increase (double).

An ideal battery is hooked to a light bulb with wires. A second identical light bulb is

connected in series with the first light bulb. After the second light bulb is connected,

the current from the battery compared to when only one bulb was connected.

Question

a) Is Higher

b) Is Lower

c) Is The Same

d) Don’t know

Since the total resistance goes up

by two, the current must go down by

two. Therefore lower.


connected in parallel to the first light bulb. After the second light bulb is connected,

the power output from the battery (compared to when only one bulb was connected)

Question

a) Is four times higher

b) Is twice as high

c) Is the same

d) Is half as much

e) Is one quarter as much

f) Don’t know

Since the current goes up by two.

Therefore the Power goes up by two

2

2

P IV

I R

V

R

24/08/2010

4


connected in series with the first light bulb. After the second light bulb is connected,

the light from the first bulb (compared to when only one bulb was connected)

Question

a) is four times as bright

b) is twice as bright

c) is the same

d) is half as bright

e) is one quarter as bright

Since the voltage goes down by a

factor of two, the power goes down

by 4. Therefore dimmer.

2

2

P IV

I R

V

R

What is the voltage drop across the 12 ohm resistor in the portion of the

circuit shown?

Question

a) 24V

b) 36V

c) 48V

d) 72V

e) 144V

1

12

1 1 1

8 4 8

R

1 2

1 1 1

PR R R

V IR

8Ω

8Ω

2Ω

12Ω

2 2 2 4R

4Ω

12A

Since the top branch has 4Ω of

resistance and the bottom

branch has 12Ω of resistance,

therefore 3 times as much

current will flow through the 4Ω

resistor than the 12Ω resistor.

This breaks down the 12A into

9A up and 3A down. So form

V=IR we have V=(3A)(12Ω)=36V

We could also have produced a RT and found 3Ω value, therefore a

voltage drop of (3Ω)(12A)=36V on both the upper and lower branch

What is the current through the 8Ω resistor in the circuit shown?

Question

a) 0.5A

b) 1.0A

c) 1.25A

d) 1.5A

e) 3.0A

V IR

2Ω

8Ω

2Ω2Ω

Since points a and b are grounded,

they are all at the same potential

(call it zero). Travelling from b to a

across the battery , the potential

increases by 24V, therefore it must

decrease by 24V across the 8Ω

resistor as we reach point a.

Therefore I=V/R =(24V)/(8Ω)=3A.

2Ω 2Ω

2Ω 2Ω

2Ω24V

a b

Question

How much energy is dissipated as heat in 20 seconds by a 100 Ω resistor that

carries a current of 0.5A?

a) 50 J

b) 100 J

c) 250 J

d) 500 J

e) 1000J

E Pt

2P I R

2

2

0.5 100

25

25

P

s

R

J

I

A

W

25 20

500

Js

s

E Pt

J

24/08/2010

5

What is the time constant for the circuit shown?

Question

a) 0.01 s

b) 0.025 s

c) 0.04 s

d) 0.05 s

e) 0.1 s

50 200

250

TR

RC

200Ω

50V

4250 2 10

0.05

F

R

s

C

200 uF50Ω

A 100Ω, 120Ω, and 150Ω resistor are connected to a 9-V battery as in the

circuit shown below. Which of the three resistors dissipates the most

power?

Question

a) The 100Ω resistor

b) The 120Ω resistor

c) The 150Ω resistor

d) Both the 120Ω and the 150Ω

e) All dissipate the same power 2

2

1

5.4

100

0.29

VP

R

V

W

100Ω

120Ω9V

150Ω

22 V

P IV I RR

2

2

2

3.6

120

0.108

VP

R

V

W

2

2

3

3.6

150

0.086

VP

R

V

W

Therefore the 100Ω

resistor dissipates the

most amount of power.

We can also answer this question faster

by noticing that the voltage drops across

the 100Ω and then across the parallel

resistors. Since the 100Ω has a higher

value than the parallel combination, it will

have a larger voltage drop than the

combination so via P=IV, it dissipates the

most power.

Question

A 1.0 F capacitor is connected to a 12 V power supply until it is fully

charged. The capacitor is then disconnected from the power supply, and

then is used to power a toy car. The average drag force on the car is 2 N.

about how far will the car go?

a) 36m

b) 72m

c) 144m

d) 24m

e) 12m

21

2CU CV

W Fd

2

211.0 12

2

1

72

2C

F

C

V

U V

J

72 2

36

J N d

d

W d

m

F

First we find the

energy stored in

the capacitor

This energy is the Work

used in moving the car a

fixed distance.

Question

Three capacitors are connected to a 9 V power supply as shown. How

much charge is stored by this system of capacitors?

a) 3 uC

b) 30 uC

c) 2.7 uC

d) 27 uC

e) 10 uC

C1=2uF

C2=4uF

C3=6uF

9V

1

1

1 1 1...

...

S n

P n

C C C

C C C

Simplify the circuit to find

equivalent capacitance.

1 2

2 4

6

EQ

F F

F

C C C

3

1 1

6

1

3

1 1

6

T

T EQ

F F

C

C F

C C

To determine the charge , we use Q=CV

3 9

27

F

C

C

Q V

V

Q CV

24/08/2010

6

Question

What is the resistance of an ideal ammeter and an ideal voltmeter?

a) zero

b) infinite

c) zero

d) infinite

e) 1Ω

a) infinite

b) zero

c) zero

d) infinite

e) 1Ω

Ideal Ammeter Ideal Voltmeter

Measures current Measures Voltage

An ammeter is placed in series with

other circuit components. In order

for the ammeter not to itself resist

current and change the total current

in the circuit, you want it to have

zero resistance.

A voltmeter is placed in parallel

with other circuit components. If it

had a low resistance, the current

will flow through it instead of the

other circuit element. So you want

it to have infinite resistance to it

won’t affect the circuit element

being measured.

Question

A light bulb is rated at 100 W in Canada, where the standard wall outlet

voltage is 120V. If this bulb was plugged in in england, where standard wall

outlet voltage is 240V, which of the following would be true?

a) The bulb would be ¼ as bright.

b) The bulb would be ½ as bright.

c) The bulb’s brightness would the same.

d) The bulb would be trice as bright.

e) The bulb would be 4 times as bright

Your first instinct is to say that because brightness depends on power, the bulb

is exactly as bright. But that is not correct. The power of the bulb can change.

22 V

P IV I RR

The resistance of a light bulb is a property of the bulb itself, and so will not

change no matter what the bulb is hooked to.

Question

A current of 6.4A flows in a segment of copper wire. The number of electrons

crossing a cross-sectional area of the wire every second is about?

a) 6.4

b) 4 x 1019

c) 4 x 10-19

d) 6.4 x 1019

e) 6.4 x 10-19

qI

t

Since each electron carries a charge of 1.6 x 10-19 C. The we calculate that

1

19

91.6 10

4 10

6.41

qI

t

N CA

s

N

Question

In a house trailer there is a 120V battery available for use. A toaster is designed to

work properly as 120V, where it is rated at 1200W, and a 120W blender which is

only designed to work properly at 60V. You look around the trailer and find a

supply of 60Ω resistors

a) What is the resistance of the toaster and blender at their rated

voltages?

b) Determine the current that is achieved in the blender when it is

working properly.

c) Create a circuit that will make both devices work simultaneously.

d) What power must the battery supply to run your circuit?

24/08/2010

7

Question





a) What is the resistance of the toaster and blender at their rated voltages?

Since we know the Power requirements , let’s us a power formula:

2VP

R

2

2120

1200

12

Toastertoaster

Toaster

VR

P

V

W

2

260

120

30

blenderblender

blender

VR

P

V

W

Question





b) Determine the current that is achieved in the blender when it is working

properly.

This is a job for Ohm’s Law: V IR

3

2

60

0

VI

R

V

A

Question





c) Create a circuit that will make both devices work simultaneously.

60Ω

blender

Toaster

ε=120V

60Ω

Required to give a

current of 2A for

blender

Question





d) What power must the battery supply to run your circuit?

V IR

120

12

10

VI

R

V

A

From Ohm’s Law, we can determine current in toaster

Since the blender had a current of 2A, this gives

the circuit a total current of 12A

Using P=IV, we have:

12 12

1440

0

P IV

A V

W

24/08/2010

8

Question

Given the following circuit

a) At what rate does the battery deliver energy to the circuit?

b) Determine the current through the 20 Ω resistor.

c) i) Determine the potential difference between points a and b

ii) At which of these two points is the potential higher?

d) Determine the energy dissipated by the 100 Ω resistor in 10 s

e) Given that the 100 Ω resistor is a solid cylinder that’s 4 cm

long, composed of a material whose resistivity is 0. 45 Ωm,

determine its radius.

10Ω

10Ω

20Ω

100Ω

40Ω

10Ω

ε=120V

a

b

QuestionGiven the following circuit

a) At what rate does the battery deliver energy to the circuit?

10Ω

10Ω

20Ω

100Ω

40Ω

10Ω

ε=120V

Recall: Rate is Power

P IV

We need to write this as

a simple circuit with a

RT so that we can

determine the current, I

by using V=IR.

110 10 10

1 1

40

60

20 100

TR

1202

60

VI

V

RA

2 120

240

A V

V

W

P I

RT


b) Determine the current through the 20 Ω resistor.

10Ω

10Ω

20Ω

100Ω

40Ω

10Ω

ε=120V

Recall: from

Question a) we

have a 2A current

2A

2A

The ratio of the

resistance of left

side to right side is

40:120 or 1:3

Therefore ¼ of 2 A goes through the right and ¾ of 2A goes through the left.

Thus (1/4)(2A)=0.5A passes through the 20Ω resistor.

120V-

(2A)(10Ω)-

(2A)(10Ω)-

(2A)(10Ω)

=60V

We can

determine the

voltage drop .

Then we use

V=IR to find

each individual

current.

OR


c) i) Determine the potential difference between points a and b

ii) At which of these two points is the potential higher?

10Ω

10Ω

20Ω

100Ω

40Ω

10Ω

ε=120V

a

b

i)

20 100

0.5 20 0.5 1

6

00

0

a bV V IR IR

A A

V

ii) Point a is at a higher

potential. Since current

flows from a high potential

to a low potential.

24/08/2010

9


d) Determine the energy dissipated by the 100 Ω resistor in 10 s

10Ω

10Ω

20Ω

100Ω

40Ω

10Ω

ε=120V

a

b

Recall: Energy equals Power

multiplied by Time

E Pt

2P I R

2

20.5 100 10

250

E I Rt

A s

J

Question

Given the following circuit

e) Given that the 100 Ω resistor is a solid cylinder that’s 4 cm

long, composed of a material whose resistivity is 0. 45 Ωm,

determine its radius.

20Ω

100Ω

10Ω

10Ω

40Ω

10Ω

ε=120V

a

b

BB

B

LR

A

2A r

2

BB

LR

r

B

B

Lr

R

3

0.45 0.04

1

7.6 10

00

mr

m

m

QuestionGiven the following circuit with the switch S turned to point a at time t=0.

Express all answers in terms of C, r, R, ε, and constants.

a) Determine the current through r at time t=0.

b) Compute the time required for the charge on the capacitor to reach

one-half its final value.

c) When the capacitor is fully charged, which plate is positive?

d) Determine the electric potential energy stored in the capacitor when

the current r is zero.

When the current through r is zero, the switch S is moved to b [t=0].

e) Determine the current through R as a function of time.

f) Find the power dissipated in R as a function of time.

10Ω

RC

r

ε

S

ba



a) Determine the current through r at time t=0.

10Ω

RC

r

ε

S

baV IR

10I

r

24/08/2010

10



b) Compute the time required for the charge on the capacitor to

reach one-half its final value.

10Ω

RC

r

ε

S

ba 1t

rCiQ t Q e

10

10

11

2

1

2

1ln

10 2

10 ln 2

t

r C

t

r C

e

e

t

r C

t r C

Therefore we need



c) When the capacitor is fully charged, which plate is positive?

10Ω

RC

r

ε

S

ba

Because the top plate is connected to the positive end of the battery,

the top is positive.

+



d) Determine the electric potential energy stored in the capacitor

when the current r is zero.

10Ω

RC

r

ε

S

ba21

2U CV

When the current through r is zero, the capacitor is fully charged, with

voltage across the plates matching the emf of the battery.

21

2U C




e) Determine the current through R as a function of time.

10Ω

RC

r

ε

S

ba

0

t

RCI t I e

The current established by the discharging capacitor decreases

exponentially.

0

t

RC

t

RC

I t I e

eR

24/08/2010

11




f) Find the power dissipated in R as a function of time.

10Ω

RC

r

ε

S

ba2P I R

0

t

I t I e

2

2

22

t

t

RC

P I t R

e RR

eR

RC

Question

a) Simplify the above circuit so that it consists of one equivalent resistor

and the battery.

b) What is the total current through this circuit?

c) Find the voltage across each resistor.

d) Find the current through each resistor.

e) The 500Ω resistor is now removed from the circuit. State whether the

current through the 200Ω resistor would increase, decrease, or

remain the same.

200Ω

ε= 9V

300Ω

400Ω

500Ω

Question

a) Simplify the above circuit so that it consists of one equivalent resistor

and the battery.

200Ω

ε= 9V

300Ω

400Ω

500Ω

1 1 1

400 500

2000

9

222.2

R

R

1 1 1

200 300

120

R

R

2000120

9

342.2

EQ

EQ

R

R

REQ

Question

b) What is the total current through this circuit?

200Ω

ε= 9V

300Ω

400Ω

500Ω

9

342.2

0.0263

V IR

VI

R

V

A

REQ =342.2Ω

24/08/2010

12

Question

c) Find the voltage across each resistor.

d) Find the current through each resistor.

200Ω

ε= 9V

300Ω

400Ω

500ΩV I R

200Ω 200Ω

300Ω 300Ω

400Ω 400Ω

500Ω 500Ω

Total 9V 0.0263A 342.2Ω

1 1 1

200 300

120

R

R

0.0263 120

3.16

V IR

A

V

9 3.16

5.84

V V V

V

Let’s use a VIR chart

3.16V

3.16V

0.0158A

0.0105A

5.84V

5.84V

0.0146A

0.0117A

Question

e) The 500Ω resistor is now

removed from the circuit. State

whether the current through the

200Ω resistor would increase,

decrease, or remain the same.

200Ω

ε= 9V

300Ω

400Ω

500Ω

By removing a resistor

from a parallel set, we

actually increase the

resistance of the total

circuit.

Therefore by Ohm’s

law if the voltage

remains the same and

the resistance

increases, the total

current must decrease

Now through the 200Ω set, the

total resistance remains the

same, yet the current decreases,

therefore the voltage across each

resister decreases as well as the

current .

circuits - the burns home page practice problems ppt.pdfthree resistors are connected to a 10-v ......

Documents