UCONN Physics 1 & 2 (Scanlon) Chapter 23 Notes: Circuits

What is Ohm's Law?

What is Ohm's Law?

I = V/R

What does a battery look like?

What does a resistor look like?

What does a junction look like?

What does a capacitor look like?

(sides of equal length; unequal for battery)

What does a switch look like?

Junction Rule

current in = current out

Answer the following:
. a wire of 5 A splits into two smaller wires, where one has 2A. How many A does the other wire have?
. Two wires of 4A and 7 A respectively combine at a junction. How many A does the larger wire have?

. 3 A
. 11 A

Loop Rule

in any loop the sum of the ∆V potential differences has to be equal to 0

How does the Loop Rule work?

- pick a point on the circuit
- go around the look measuring the gains or losses in potential in both directions

Apply the Loop Rule to the following and find its current:
a battery with 12 V on one end and 0 V on the other. The resistor has 6 ohms

current: I = ∆V/I = 12 V/6 ohms = 2A
εV = +12 V - 12 V = 0

A circuit has a potential of 12 V and two resistors of 7 ohms and 5 ohms respectively. Apply Loop Rule.

εV = +12 V - 7 V - 5 V = 0
εV = + 5 V + 7 V - 12 V = 0

series circuit

a circuit with one path and no junctions
ex. 1 battery of 6V, one resistor of 12 Ω

Describe a series circuit with a battery of 6V and resistance of 12 Ω. What is the current?

the current, which is constant throughout the circuit, is 1/2 A
I = V/R = 6/12 = .5 A

What happens to current if there are multiple resistors in a circuit?

the current is still the same throughout the circuit but there is less current in general

There is a circuit with a battery of 6V and four resistors with respective charges of 2Ω, 4Ω, 6Ω, and 12Ω. Find the ∆V, I, and P for each resistor.

εΩ = 24 Ω
ε∆V = 0 = + 6V - V(2Ω) - V(4Ω) - V(6Ω) - V(12Ω)
so, V(2Ω + 4Ω + 6Ω + 12Ω) = 6 V
6V = I₂R₂ + I₄R₄ + I₆R₆ + I₁₂R₁₂
if they all have the same current, then I₂ = I₄ = I₆ =I₁₂
6V = I₂R₂ + I₄R₄ + I₆R₆ + I₁₂R₁₂
6V = I(24Ω) so I = 0.25 A
P = IV so multiply .25A by each V to find it at each point; these add up to 6V
Total power: P = (.25)(6V) = 3/2 W

What should you assume for parallel circuits?

that the wires are ideal and there is no energy lost

Describe parallel circuits:
- current
- potential difference/voltage

- different currents
- all elements see the same potential difference/voltage

A parallel circuit has a battery with a ∆V of 12 V. What happens if you attach a voltmeter to either side of each resistor in parallel circuits?

one side of the voltmeter is on the 12 V side (upper half if split horizontally) and one is on the 0 V side (lower half) so the ∆V still = 12 V

A series circuit has a 12 V battery and two resistors with a resistance of 4Ω and 2Ω. What is the R(EQ) and the I?

R(EQ) = R(1) + R(2) = 4Ω + 2Ω = 6Ω
I = V/R = 12 V/2Ω = 2 A

What is current equal to for parallel circuits?

I = I(1) + I(2) + I(3)

You have a series circuit with ∆V = V and R1, R2, and R3. How does I move? How can you find I?

I moves around the circuit from V in a clockwise manner and down the parallel parts of the circuit from top to bottom. Depending on where you are in the circuit, you can add or subtract I (ex. I2 + I3 for a middle circuit box)
V(battery)/R(EQ) = V1/R1 + V2/R2 + V3/R3
V total = V1 = V2 = V3
I = V(1/R1 + 1/R2 + 1/R3)
**switch REQ so no longer inverse

How can you find REQ for a circuit?

1/REQ = 1/R1 + 1/R2 + 1/R3

A series circuit has a 12 V battery and a current of 12 A initially. The R1 = 2 ohms, R2, = 4 ohms, and R3 = 6 ohms. Find I1, I2, I3, and I4. Then, find REQ.

I1 = 12 V/2 ohms = 6A
I2: 11 A - 6A = 5A. 12 V/4 ohms = 3A
I3: 5 A - 3 A = 2A
1/REQ = 1/2Ω + 1/4Ω + 1/6Ω
= 6 A/12 V + 3A/12 V + 2A/12 V = 11/12 Ω
find inverse; REQ = 12/11 Ω or 1.09 Ω
12 V/1.09 A = 11 A

What is the purpose of a circuit breaker? Where is it placed?

it is protection against too much current and prevents fires; it is placed where current is the greatest

Describe a vacuum cleaner

when there is more current then it knows what to do with the breaker shuts off

What happens to resistance when more appliances are added?

it increases

What should REQ be smaller than?

the smallest R

What happens as R decreases?

I increases

There is a parallel circuit with a current of 4 A and a V of 120 V. Find the current if the power is 100 W. What is the current if the power is 40 W and how does this differ?

P = IV
100 = I(120) = 0.8 A
40 = I(120) = 0.33 A
when resistance increases, less current is drawn

Describe the relationship between P and R and I and R

when P increases, R decreases (inverse relationship)
When R decreases, I increases (inverse relationship)

rheostat

adjustable resistor

Describe R for the following:
- fast fan
- slow fan

- low R
- high R

There is a series circuit with a V of 12 V and four resistors: 4 Ω and 6Ω in a series, and 6Ω and 8 Ω in a voltage trap.
- Find REQ
- Find A
- V for each
- Which bulb is the brightest?

- for the voltage trap, 1/REQ = 1/6Ω + 1/8Ω = 13.43 Ω
- 12 V/13.43 Ω = 0.89 A
- V = RI so R(.89). At 4 Ω V = 3.56, at 6Ω = 5.34. For the voltage trap 12 V - 3.56 - 5.34 = 3.1 V
3.1 V/6Ω = .51 A; 3.1V/8Ω = .379 A
*more current through the 6Ω resistor
- 6Ω (largest V drop and current), 4Ω, 6Ω in voltage trap, 8Ω

How can you find which bulb is the brightest?

look for greatest power consumption (largest V drop and current)

A series circuit has a battery of 12 V with 4Ω (3.56 V) and 6Ω (5.34 V) resistors. It is also hooked to a miniature parallel circuit with 6Ω (3.1 V) and 8Ω resistors (3.1 V)
- does the loop have to have a battery?
- prove that Loop Rule applies to both loops
- what happens in the opposite direction for Loop Rule?

- no it does not
- ε∆V = 0;
for the series loop + 12 - 3.56 - 3.1 - 5.34 = 0
for the parallel loop + 3.1 - 3.1 = 0
- the signs are reversed

Simplify Junction Rule

current in = current out

Describe a series circuit

- one path (no junctions)
- current (I) is the same no matter how many resistors there are
- R(EQ) = R1 + R2 + R3
- use V = IR

Describe a parallel circuit

- junctions (current splits and rejoins)
- all elements and resistors have the same ∆V
- 1/REQ = 1/R1 + 1/R2 + 1/R3
- use V = RI

What does a battery force electrons to do?

be on the move

Describe capacitors in series

- use Q = C∆V instead of V = RI
- Q is the same for all capacitors (Q1 = Q2 = Q3)
- current is not always flowing in the circuit (ex. not once the capacitors are charged)
- 1/CEQ = 1/C1 + 1/C2 + 1/C3
- V(total) = V1 + V2 + V3
- battery only charges one capacitor

How can you find V using Q = CV using V in capacitors in series?

V = Q/C

A series circuit has a battery of 12 V and three capacitors. How are electrons distributed?

the battery charges on plate on one capacitor positively and one plate on another capacitor negatively. Electrons and protons redistribute from capacitor to capacitor, and the middle capacitor is charged by induction

A series circuit has a batter of 12 V and 3 capacitors. C1 = 2 µF, C2 = 4 µF, C3 = 8 µF.
Find Q, CEQ, and V for each

1/CEQ = 1/2 + 1/4 + 1/8 = 7/8 = 8/7 = 1.14 µF
Q = CV = 12 V * 1.14 µF = 13.68 µC
V for C1 = 6.84 V (13.68/2), V for C2 = 3.42 V, V for C3 = 1.71 V

Describe the current (I) on capacitors when fully charged

I = 0; current is done

capacitance efficiency

how much charge (Q) is stored v. ∆V

Describe capacitors in parallel:

- Q total = Q1 + Q2 + Q3
- CEQ = C1 + C2 + C3
- V1 = V2 = V3 (all the same V)
- battery charges all capacitors

What should you do with both CEQ and REQ?

draw a new graph with V (battery) on one side and REQ/CEQ on the other

Describe:
- REQ/series
- REQ/parallel
- CEQ/series
- CEQ/parallel

- REQ = R1 + R2 + R3
- 1/REQ = 1/R1 + 1/R2 + 1/R3
- 1/CEQ = 1/C1 + 1/C2 + 1/C3
- CEQ = C1 + C2 + C3

A parallel circuit with capacitors has a battery of 12 V. C1 = 2 µF, C2 = 4 µF, and C3 = 8µF. What is Q(total) and what is CEQ?

CEQ = 2 + 4 + 8 = 14 µF
Q(total) = CV = (14)(12) = 168 µC

A series circuit has a battery of 12 V and the capacitors 12 µF and 2 µF. There are two linked parallel circuits attached to it: one with 4 µF and 6 µF, and one with 8µF and 10 µF. Find the CEQs of both parallel circuits, the voltages of each capacitor, and the charges associated with each capacitor.

CEQ = 10 µF for the first parallel circuit and 18 µF for the second circuit
total CEQ = 1/CEQ = 1/2 + 1/10 + 1/12 + 1/18 = 1.35 µF
Q = 12 V * 1.35 µF = 13.35 µC everywhere
V for 2 µF capacitor = 13.35/2 = 6.68 V; V for 12 µF capacitor = 1.11 V
Parallel #1: 13.35/10 µC = 1.34 V; 1.34 * 4µ F = 5.36 µC; 1.34 * 6 µF = 8.04 µC
Parallel #2: 13.35/18 µC = 0.74 V; .74 * 8µF = 5.9 µC; .74 * 1- µF = 7.4 µC

Describe RC circuits

fully charged capacitor + a resistor + a switch so it's an open circuit. I flows clockwise, e- flow counterclockwise

What does a capacitor in an RC circuit do?

it stores charge
ex. in a car it makes the windshield wipers move

Describe Vc on a loop

Vc = ε∆V
Vc changes as charged leaves because VR gets smaller

How can you find I in an RC circuit?

I = I₀e^⁻(t/RC)
where Io = the initial current in a circuit
exponential decay

What is tau?

resistance * capacitance (RC)
time constant; units = s

How can you find I₀?

I₀ = Vc/R

On a graph of I v. t, what happens when t = 0?

the switch is closed

What happens if t = RC on a graph of I v. t?

I₀e⁻¹ = 0.37 I₀
I₀e⁻² = 0.135 I₀

How can you find voltage on an RC circuit?

V = Vc*e^⁻(t/RC)
exponential decay

Describe discharge for the following:
- large resistance
- small resistance

- large discharge
- low discharge

What happens to R on a parallel circuit with a battery on one wire, a capacitor on another, a resistor on the other, and open switches on the two horizontal sections?

I very rapidly flies off the circuit and goes to 0

What is another formula for Vc?

Vc = E₀(1 - e^-t/RC)
Vc = E₀ - E₀e^⁻t/RC
exponential gain, same for both discharge and charging

What happens when C or V are at 1% of their original value?

it still has 99% percent of its charge

An RC circuit has 8 µF and 500 Ω. What is RC?

RC = (8*10⁻⁶ F)(500 Ω)
= 0.004 s

An RC circuit has 8 µF, 500 Ω, and 12 V.
a) Describe I and find I₀.
b) 0.005 seconds after you close the switch, how much current is in the circuit?
c) Find Vc

a) I is fully charged so it can only discharge
I₀ = V/R = 12 V/500 Ω = 0.024 A
b) I = I₀e^-t/RC
I = 0.024*e^-(.005/.004) = 0.00688 A
c) Vc = (12 V)*e^-(.005/.004)

An RC circuit has 8 µF and 500 Ω; V is 12 V and Vc is 4 V. Find the time when Vc = 4V-how long after the switch closes?

Vc = E₀(1 - e^-t/RC)
4V = 12V(1 - e^-t/.004)
-8V = -12V*e^-t/.004
0.667 = e^-t/.004
apply ln to both sides;
ln (.667) = ln e^-t/.004
-0.405 = -t/.004
t = 0.00162 s

A single resistor is wired to a battery as shown in (Figure 1) below. Define the total power dissipated by this circuit as P(0) if the second resistor is now removed from the circuit and rewired in parallel with the original resistor.

This turns into a circuit with voltage V and two resistors in parallel; if one resistor in series with the battery is P(0), two parallel circuits = 2P(0)

You have been given a long length of wire. You measure the resistance of the wire, and find it to be R(long). You then cut the wire into n identical pieces. If you connect the n pieces in parallel, what is the total resistance R(total) of the n wires connected in paralle?

R = l/A
R(short) = R(long)/n
n/[R(long)/n) = n²/R(long
REQ = 1/REQ = R(long)/n²

You are given a series circuit with a battery of voltage V and four wires: R1 with area A, R2 with area 2A, R3 with area 3A, and R4 with area 4A. Find the voltage across wire 2.

R2 = R/2, R3 = R/3, R4 = R/4
connected in series so
REQ = R1 + R1/2 + R1/3 + R1/4 = 25(R1)/12
current for each = I = Vo/R
Vo = IR = I(R2) = I(R1)/2
= 6Vo/25 = 0.24 Vo

In the circuit in (Figure 1), a 20-ohm resistor sits inside 110 g of pure water that is surrounded by insulating Styrofoam. If the water is initially at temperature 10.3 ∘C, how long will it take for its temperature to rise to 57.0 ∘C? Use 4190 J/kg⋅∘C as the specific heat of water, and express your answer in seconds using three significant figures.

REQ = 30 ohms; V = 30 V so I = 1A
P = I²R = (1)²(20Ω) = 20 W)
Q = mC∆T = P∆t
(.11 kg)(4190 J/kg*C)(46.7°C) = (20 W)∆t
∆t = 1080 s

An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru (Figure 1). The ammeter reads a current I0. Next, a resistor of unknown resistance Rr is connected in series to the ammeter, and the ammeter's reading drops to I1. Finally, a second resistor, also of resistance Rr, is connected in series as well. Now the ammeter reads I2.

Io = Vb/Ru I1 = Vb/(Ru+Rr)
IoRu = I1(Ru + Rr)
4/5 = Ru/(Ru+Rr)
Ru = 4 Rr; Rr = 1/4Ru
R(total) = Ru + .5Ru = 3/2 Ru
I2/Io = Ru/1.5 Ru = 2/3 = 0.667

When switch S in the figure is open, the voltmeter V of the battery reads 3.08 V . When the switch is closed, the voltmeter reading drops to 2.91 V , and the ammeter A reads 1.69 A . Assume that the two meters are ideal, so they do not affect the circuit.
a) What is the emf?
b) Find the internal resistance r of the battery
c) Find the circuit resistance R

a) 3.08 V
b) ∆V = 3.08 - 2.91 = 0.17 V/1.69 A = 0.1006 Ω
c) 2.91 V/1.69 A = 1.72 Ω

Three capacitors are connected to each other in series, and then to the battery. The values of the capacitances are C, 2C, and 3C , and the applied voltage is ΔV. Initially, all of the capacitors are completely discharged; after the battery is connected, the charge on plate 1 is Q.
c) find the voltage ∆V1 across the first capacitor
e) Using the value of Q just calculated, find the equivalent capacitance Ceq for this combination of capacitors in series.

c) V = V1 + V2 + V3
V1 = Q/C, V2 = Q/2C, V3 = Q.3C
= 6/6 + 3Q/C + 2Q/6 = 11Q/C
V(total) = 11Q/6C,, V1 = 6Q/6C
V1/Vtotal = (6Q/6C)/(11Q/6C) = 6∆V/11
d) 1/CEQ = 1/C + 1/2C + 1/3C
= 6/6C + 3/6C + 2/6C = 11/6C

You are given two circuits. Rank the bulbs from brightest to dimmest.
a) a parallel circuit with A in one section, a battery in the middle, and B and C in series in the second section
b) A series circuit has one bulb A and a battery, which are connected to an internal circuit with B and C in parallel

a) A is the brightest because it doesn't share power and B and C are equally dim
b) bulb A is the brightest because it doesn't share current and B and C are equally dim; they have the same power but it is lesser

There is a parallel circuit with a battery in one section, a resistor in the middle section, and a resistor in the third section with an open switch above it. Does the current through the battery increase, decrease, or stay the same when the switch is closed?

when you close the switch there is more current; I = V/R. V stays the same and R decreases because REQ = 1/R1 + 1/R2
= 2x as much current as when closed

You are given a series circuit with a 9 V battery and a 5 ohm and 10 ohm resistor respectively. An ammeter is attached to either end of the 5 ohm resistor (below the 10 ohm resistor). What is the current through it?

A is nearly zero because current takes the path of least resistance; there is more current by the 10 ohm resistor

Three capacitors are arranged so that C2 and C3 are parallel to each other in a smaller circuit, which is connected in series to C1 with a voltage source connected across the combination. C 1 has a capacitance of 9.0 pF, C2 has a capacitance of 18.0 pF, and C3 has a capacitance of 27.0 pF. Find the potential drop across the entire arrangement if the potential drop across C 2 is 257.0 V.

C2 and C3 are parallel so they both have 257 V across them. Once CEQ and C1 are in series, they will have the same Q.
C2: Q = C∆V = (18µF)(257 V) = 4.63 µC
C3: 18 * 1.5 = 27 so 4.63 * 1.5 = 6.93 µC
Q(total) = 11.96 µC so C1 also has 11.96 µC
V = Q/C = 11.96 µC/9*10-9 F = 1284 V
∆V = 1284 V + 257 V = ~1500 V

The network shown is assembled with uncharged capacitors X , Y, and Z, with CX = 3.0 μF, CY = 5.0 μF and CZ = 1.0 μF. The battery is in parallel to CX, which is in parallel to CY and CZ (which are in series together). The switches S1 and S2 are initially open, and a potential difference Vab = 120 V is applied between points a and b. After the network is assembled, switch S1 is then closed, but switch S2 is kept open. What is the final potential difference across capacitor Z?

The voltage across CX is 120 V, so the voltage across CY and CZ in combination is also 120 V
1/CEQ = 1/1 + 1/5 = 6/5 = 5/6
Q = CV for series: (5/6µF)(120 V) = 100 µC
both CY and CZ have 100 µC of charge
Q = CV for CZ; 100 µC = (1 µF)V = 100 V

A 3.0-Ω resistor is connected in parallel with a 6.0-Ω resistor. This combination is then connected in series with a 4.0-Ω resistor. The resistors are connected across an ideal 12-volt battery. How much power is dissipated in the 3.0-Ω resistor?

REQ = 1/6 + 1/3 ohms = 3/6 ohms = 2 ohms
2 ohms + 4 ohms; R(total) = 6 ohms
I = V/R = 12 V/6 ohms = 2 A
V = IR = (2 A)(4 Ω) = 8V
4 V drop over 3 ohm resistor
P = V²/R = (4)²/3Ω = 5.3 W

A battery V has R1 (10 Ω) and R4 (30 Ω) connected to it in series. These are connected to R2 and R3 in parallel, which both have 20 Ω. If 1.5 A flows through R2, what is the emf V of the ideal battery in the figure?

R2 = R3 bc parallel, 1.5 A through both parallel circuits. V = IR = (1.5 A)(20Ω) = 30 V for each
current flowing out of parallel circuit towards R4 = 3 A; V = IR = (3 A)(30 Ω) = 90 V
same with R1; V = IR = (3A)(10 Ω) = 30 V
ignore second parallel circuit and form one series mentally; 30 V + 30 V + 90 V = 150 V

A battery of 9 V is directly opposite a battery of 6 V and both are connected in series to a resistor of 12 Ω. The square this circuit forms is split diagonally starting at the top left (6 V on top) and ending on the bottom right (9 V on bottom) with the 12 Ω resistor to the left. The diagonal wire has a resistor of 18Ω.
Determine the current in the 12-Ω resistor for the circuit shown in the figure assuming that the batteries are ideal.

Redraw it so that the 6 V battery, 18 Ω resistor, and 9 V battery are in parallel (switch from diagonal to horizontal line). The 6V battery is connected in series to the 12 ohm resistor on the top layer, then the 18 Ω resistor in the middle, then the 9 V resistor on the bottom.
Loop Rule starting on bottom: +9 V - 6 V - V(12 Ω resistor = 0
so 3 V voltage drop for 12 Ω resistor
I = V/R = 3V/12Ω = 0.25 A

There is a circuit with a 12 V battery, a 5 ohm resistor in series, and a parallel circuit in series with it. On one side of the parallel circuit is an 8 ohm resistor, and on the other side a 2 ohm resistor and a 4 ohm resistor in series.
d) Describe what would happen to the voltage, current, and power of each bulb (increase/decrease/stay the same) if the 4 ohm bulb burned out causing no current to pass through the bulb.

4 ohm bulb: goes out so nothing happens
2 ohm bulb is also out so nothing happens
5 ohm bulb sees all current so V, I, and P decrease (dimmer)
8 ohm bulb; I increases, V increases, P increases = brighter