Chapter 20 - Electric Circuits Flashcards
A tightly coiled spring having 75 coils, each 3.50 cm in diameter, is made of insulated metal wire 3.25 mm
in diameter. An ohmmeter connected across opposite ends of the spring reads 1.74 Ω. What is the resistivity
of the metal?
1) Find total length of wire
Calculate the circumferance of each loop
πd = π(0.0350m)
or 2πr
Multiply this by 75
2) Find cross sectional area of wire
Wire has radius of diameter/2
0.0035/2
A = πr^2
=8.29x10^-6
Input the calculated values into the resistivity equation
Resistor 1 and 2 in parallel and 3 and 4 in parallel and 1&2 in series with 3&4. Explain how to caluate the current in each resistor given resistance in each resistor?
1) Calculate equivalent resistance
2) Calculate current in entire circuit
3) Calculate voltage in each pair of circuits (since current is the same for circuits in series)
4) Then calculate the current in each resistor knowing this voltage and the resistance
In what direction does the conventional current flow in?
From high to low potential
Answer with whether each of the following combinations has same/differnet charge and voltage
Resistors in series
Capacitors in series
Resistors in parallel
Capacitors in parallel
Resistors in series → Same current, different voltage
Capacitors in series → Same charge, different voltage
The reason for this makes sense and is as such:
Current = q/t so charge follows suit with the rule of current
Resistors in parallel → Same voltage, different current
Capacitors in parallel → Same voltage, different charge
In a circuit, a 3F (C1) and 5F (C2)capacitor are in parallel with each other and in series with a 6F (C3) capacitor. The potential difference between the two is 24V.
a) Calculate the charge and potential difference across each capacitor
1) Calculate the Ceq
Since 3 and 5 are in parallel their equivalent is 8F and it’s in series with 6F so the equivalent is 3.43 F
2) Calculate the total charge
Q = CV
Q = 8.23 x 10^-5
Charge is the same for circuits in series, so the C1C2 = C3 and C3 has a charge described above
Now find the voltage across each of the circuits since voltage is shared in parallel
V12 = Q12/C12
= 10.3 V
So the charge would be Q1 = V1C1 (10.3 x 3) = 3.09 x 10^-5
Q2 = (10.3)(5 x 10^-6) = 5.15 x 10^-5
With the capcitance of each it is pretty easy to find the potenital difference of each
V1 = 10.3
V2 = 10.3
V3 = 13.7
Emf with closed vs open circuit
In an open circuit, there is no current so emf equals the voltage of the battery
In closed circuit, terminal volage is
V = e - Ir so e is emf
r = internal resistance
A 12V battery has internal resistance of 1ohm and is connected in series with an external resistor 5ohms.
a) Find the rate of conversion of internal chemical energy to electrical energy within the battery
b) Find the rate of power dissipation of electrical energy in the battery
PART A
P = VI
to find I consider
V = IR
total resistance is 6 ohms and V = 12 so I = 2A
Sub back into power equation to get 24W
This represents the total pwoer
PART B
Power disapation is P = I^2r
so you will get 4W
You can use current because the resistors are in series so current is shared
____ is the same through each device in series
_____ is the same through each device in parallel
And analogies of current and voltage through a hose
Current is like rate of flow of water
Voltage is like water pressure
Resistors reduce the pressure but the flow is the same across both in series
Voltage is the same in parallel because all paths are connected to the same source so pressure same entering
a) current
b) voltage
How do you prove which end of the resistor is of higher potenital knowing current is 5A and flows from left to right
Current flows from higher to lower potential and the current (positive so correct direction) through the resistor flows form left to right, so left has to be higher and hence A
What factors affect and do not affect capacitance
a) Area of plate
b) Separation of plate
c) Material between plates
d) charge on plates
e) Energy stored in capacitor
a, b, and c yes because of the equation for capacitance but charge and energy do not affected. even tho they appear in the equation, capacitance would remain constant
The power rating of a lightbulb is the power it consumes when connected across a
120 V outlet. (a) If you put two 100W bulbs in series across a 120 V outlet, how much power would each
consume if its resistance were constant? (b) How much power does each one consume if you connect them
in parallel across a 120 V outlet?
PART A
1) Find the resistance of one lightulb
R = V^2/P
144
2) Find total resistance in series 144+144 = 288
3) Find total current (shared in series)
I = V/R
0.417A
4) Find power for one lightbulb
P = I^2R
= 0.417^2x414
= 25W
PART B)
Voltage is the same in parallel and you know the resistance of one lightbulb is 144 ohms
so the power disapated by each is 100W
EXAM QUESTION
Picture in word document
Question with capacitors and ammeter and voltmeter
a) Find the maximum charge on each capactior
b) Find the maximum potential difference on each capacitor
d) find the time constant for the circuit
a)
Find Cs (9.23) and Cp (10)
The charge is the same for capacitors in series so take equivalent capitance and find each charge using Q = CV
For the ones in parallel you get 461pC (9.23 x 50) and for the one in series you get 500 pc (10 x 9.23)
b) For the capactitor in parallel, voltage is the same (so 50V)
For the ones in series, you do V=Q/C, using the same charge for each but differing capacitances
d) T = RC where R is total risstance and C is total capacitance
TOtal resistance of circuit is 20 ohms and total cpacitance is teh capacitance in sereis calculated (9.23) and in parallel (10) added together.
so final answer is 384
. A person with a body resistance of 10 kΩ between his hands accidentally
grasps the terminals of a 14 kV power supply. (a) If the internal resistance of the power supply is 2000 Ω,
what is the current through the person’s body?
Entire voltage is supplied across both the person and the power supply, so you have to consider total resistance, otherwise you’re ignoring some of the resistance the current passes through
Hence I = V/Rtot
1.17A
How would you define electromotive force?
Maximum potential difference between battery terminals
How should ammeter and voltmeter be connected?
RC Circuti
Ammeter in series, ideal has 0 resistance because current passes through
Voltmeter in paallel, ideal has infinite resistance to not draw up the current
Combo of capacitor and resistor
An object is placed at a distance of 40𝑐𝑚 in front of a simple optical element (a thin lens/a
mirror/a spherical mirror) along its optical axis. A shrunk virtual image is created by the optical
element in front of it. (“Front” is the same side as where the object is.) The height of the virtual
image is observed to be 1/4 of the height of the object.
[ /2] 4. b) Calculate the focal length of the optical element.
a)
You know magnification is 1/4
so 1/4 = -di/do
di = -10
1/f = 1/40 + 1/-10
f = -13.33
EXAM PROBLEM
Shown in word document
R1 in series with R2, R3, R4, R5 kind of question from midterm
Find current across A to B (R3 to R4)
So R3 and R4 are in series with each other
2R
R3 and R4 are in parallel with R2 and R5
1/R = 1/2R + 1/R + 1/R
R = 2R/5
Now R1 and R2-R3-R4-R5 are in series
R + 2R/5
7R/5
P = V^2/R
60 = (58^2)/(7R/5)
R = 10.7
PART 4
Current is shared in series
So find total current
I = V/R (using total resistance value)
Check using
P = IV to ensure R value is correct
Use this to calcualte voltage in R1
V = 21.428
So 30-21.428 = 8.572 left over for parallel
Current = V/R3-4
8.572/21.4 = 0.4A
