Quiz 3 - Chapter 20 Studying 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
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
When switch S in Figure is open, the voltmeter V across the battery reads 3.08V. When the switch is closed,
the voltmeter reading drops to 2.97 V and the ammeter A reads 1.65 A. Find the emf, the internal resistance
of the battery, and the circuit resistance R. Assume that the two meters are ideal, so that they don’t affect the
circuit.
Emf is 3.08 (voltage when battery is closed)
Use V = e - Ir
isolate and solve for r using V = 2.97
To find the circuit resistance simply use V = IR
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
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
____ 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
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? (b) What is the power dissipated in his body?
a) Here you can factor in the external resistance since the current flow is the same but you don’t have just the voltage body resistance
I = V/Rtot = 14/10+2000
=1.17
have to factor in the total resistance becuase the votlage is teh total ressistance
b) For the power through him, just factor in his body resistance. And you don’t have the voltage so you will have to use
P = I^2R
13.7kJ
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
EXAM PROBLEM
In the electric circuit shown in figure, the battery has voltage of 58.0 V and power of 80 W. All the
resistances are equal. Find the value of the resistance R
The figure shows three resistors in parallel in series with two other resistors in series
First calculate the equivalent resistance
in parallel: 1/R + 1/R + 1/R = 3/R
Req = R/3
In series: R/3 + R + R = 7R/3
P = V^2/R
80 = 50^2/(7R/3)
Solve for R and you should get 18 ohms
Definition of electromotive force
Maximum potential difference between battery terminals
