Physics 150 Electric current and circuits Chapter 18 Electric current Let’s imagine a conductor with a potenDal difference between its ends -­‐ -­‐ -­‐ -­‐ NegaDve (lower) potenDal Va + + + + !
E
PosiDve (higher) potenDal Vb •  PosiDve charges will start moving to the leM (away from posiDve potenDal and toward the negaDve potenDal) •  NegaDve charges will move to the right (away from negaDve and toward the posiDve potenDal) •  Neutral atoms will just get polarized Physics 150, Prof. M. Nikolic 2 Electric current Electric current is the moDon of the charged parDcles due to the potenDal difference (voltage). Δq
I=
Δt
NotaDon for current is: I Unit of current: Ampere [A] How big is Ampere? Household ~ 1-­‐10 A
BaWery Powered ~ 100 mA
Lightning Bolt ~ 100 A to 100 kA
To hurt you ~ 40 mA
To kill you ~ 0.1 A to 0.2 A
Physics 150, Prof. M. Nikolic 3 Δq
I=
Δt
Exercise: Electric current If 5x1020 electrons flow past a given cross-­‐secDon of the metal wire in 10.0 minutes, what is the current in the wire? What do we know: Ne = 5x1020 electrons Δt = 10 min Δq
I=
Δt
First, find the total charge in the wire è Equal to number of electrons Dmes the charge of one electron Δq = Ne
Δq = 5⋅10 20 ⋅1.6 ⋅10 −19 C = 80C
Then, convert minutes to seconds: t = 10 min
Physics 150, Prof. M. Nikolic 80C
I=
= 0.13A
600
60s
= 600s
1min
4 DirecDon of the current By convenDon the direcDon of the current is equal to the direcDon of moDon of posiDve charges. (In reality, a flow of negaDve electrons determines the current). Current will always flow from high potenDal to lower potenDal -­‐ along the direcDon of the electric field. Physics 150, Prof. M. Nikolic 5 Wait, according to Chapter 16… But we stated that an electric field inside of a conductor must be zero (E=0) and a conductor is an equipoten6al surface (Vab=0). è electrons will move unDl the internal field was canceled -­‐ -­‐ -­‐ -­‐ Lower potenDal Va + + + + Higher potenDal Vb How can we achieve the non zero potenDal difference and constant movement of charged parDcles? Physics 150, Prof. M. Nikolic 6 ElectromoDve force We need an object which is capable of creaDng the potenDal difference and constantly providing electrons to one end of the conductor and removing them from the other. ε
Such an object is a source of electromo0ve force (EMF). The EMF does not create charges, but increases the energy of charges passing through it. Physics 150, Prof. M. Nikolic 7 ElectromoDve force Examples of sources of EMF are: •  Ba>eries •  Generators •  Capacitors EMF is NOT a force it is a potenDal difference (voltage). At high potenDal ε
-­‐ + At low potenDal The circuit symbol for a baWery (EMF source) The work done by an ideal baWery in pumping a charge q is W = qε. The amount of charge entering the conductor per unit Dme is equal to the amount of charge leaving the conductor per unit Dme. Physics 150, Prof. M. Nikolic 8 Microscopic view of current •  No electric field •  random moDon •  (but very fast) v ~106 m/s •  Electric field •  sDll random moDon •  with an average velocity or dri4 velocity vd~10-­‐4 m/s Physics 150, Prof. M. Nikolic 9 Microscopic view of current Physics 150, Prof. M. Nikolic 10 DriM velocity Let’s, calculate the number of charges (Ne) that pass through the shaded region in a Dme Δt. Number of charges per unit volume is Ne
Ne
n=
=
V A ⋅ Δx
N e = n(AΔx)
N e = nA(vd Δt)
Physics 150, Prof. M. Nikolic The current in the wire is: Δq eN e
I=
=
= neAvd
Δt
Δt
11 I = neAvd
Exercise: DriM velocity A copper wire of cross-­‐secDonal area 1.00 mm2 has a constant current of 2.0 A flowing along its length. What is the driM speed of the conducDon electrons? Copper has 1.10×1029 electrons/m3. What do we know: A = 1 mm2 I = 2 A n = 1.10×1029 electrons/m3 I = neAvd
I
vd =
neA
Convert 1 mm2 to m2: 1 mm2 = 1x10-­‐6 m2 vd =
29
1.10
×10
(
2.0 A
m −3 ) (1.60 ×10 −19 C) (1.00 ×10 −6 m 2 )
= 1.1×10 −4 m/sec = 0.11 mm/sec
Physics 150, Prof. M. Nikolic 12 Resistance and Ohm’s law What determines how much current we get for a given applied electric field or potenDal difference ? ➜ properDes of the conductor determine the resistance R Ohm’s law ΔV = IR
Physics 150, Prof. M. Nikolic Units of resistance: Ohm = Volt per ampere [Ω] 13 Conceptual quesDon – Ohm’s law Q1
If the potenDal difference (voltage) across a resistor is doubled: 1.
2.
3.
4.
Only the resistance is doubled Only the current is doubled Only the resistance is halved Both current and resistance are doubled ΔV = IR
Resistance is the property of the object and it does not depend on voltage and current. Physics 150, Prof. M. Nikolic 14 Resistance and resisDvity Resistance (R) is a property of an object ➜ increases with increasing length of wire – L (charge has further to “push through”) ➜ decreases with increasing cross-­‐secDonal area of wire – A (charge has more possible paths through the wire) ➜ depends upon the material the wire is made of – ρ L
R=ρ
A
Symbol for resistance: •  ρ (Greek leWer rho) is the resisDvity which is a property of the material at a given temperature. Resistance does not depend on the shape of the area A! Physics 150, Prof. M. Nikolic 15 Conceptual quesDon – Resistance Q2
A cylindrical copper rod has resistance R. It is reformed to twice its original length with no change of area. Its new resistance is: A.
B.
C.
D.
R 2R R/2 R stays the same L
R=ρ
A
What about resisDvity? ResisDvity is the property of the material it does not depend on the length and area of the resistor. Physics 150, Prof. M. Nikolic 16 Temperature dependence The resisDvity of a material depends on its temperature: ρ = ρ0 (1 + α (T − T0 ))
where ρ0 is the resisDvity at the temperature T0, and α is the temperature coefficient of resisDvity. R = R0 (1+ α (T − T0 ))
Some materials, when very cold, have a resisDvity which abruptly drops to zero. Such materials are then superconductors. Physics 150, Prof. M. Nikolic 17 Exercise: Resistance of a conductor A common flashlight bulb is rated at 0.3 A and 2.7 V. If the resistance of the bulb’s tungsten filament at room temperature (200 C) is 1 Ω, esDmate the temperature of the filament when the bulb is turned on. R = R0 (1+ α (T − T0 ))
1+ α (T − T0 ) =
R
R0
R
−1
R0
T − T0 =
α
Resistance at room temperature T0 = 200 C =20+273 K = 293 K is R0 = 1 Ω Resistance at unknown temperature T can be found from Ohm’s law ΔV = IR
R=
ΔV
I
2.7V
R=
= 9Ω
0.3A
α (tungsten) = 4.50 ×10−3 °C−1
9Ω
−1
1Ω
T = 20°C +
= 20°C+1777.8°C=1797.8°C
−3
−1
4.50 ×10 °C
Physics 150, Prof. M. Nikolic What do we know: I = 0.3 A ΔV = 2.7 V R0 = 1 Ω T0 = 200 C α = 4.5x10-­‐3 0C-­‐1 18 A simple loop circuit A current will only flow around a closed loop. ε
I=
R
Where ε is electromoDve force (EMF) The potenDal difference between the terminals of an ideal EMF device is equal to the EMF of the device Physics 150, Prof. M. Nikolic 19 A single-­‐loop rules Loop rule: The algebraic sum of all changes in potenDal encountered in a complete traversal of any loop of a circuit must be zero (Kirhhoff’s loop rule). Resistance Rule: For a move through a resistance in the direcDon of the current, the change in potenDal is -­‐iR; in the opposite direcDon +iR. EMF Rule: For a move through an ideal emf device in the direcDon of the emf arrow, the change on the potenDal is +ε; in the opposite direcDon is -­‐ε Physics 150, Prof. M. Nikolic 20 PotenDal in a circuit A liWle like water flowing downhill •  The emf is like the pump that gets the water to the top of the hill – potenDal rises •  Across every resistor potenDal drops ε − Ir − IR = 0
Physics 150, Prof. M. Nikolic 21 Ideal baWeries vs. real baWeries Ideal BaWery – Keeps potenDal difference between its terminals. Real BaWeries – have an internal resistance, r, which is small but non-­‐zero. Resul6ng (terminal) voltage supplied by the baWery ΔV = Vb −Va = ε − Ir
By applying Ohm’s law ΔV = IR = ε − Ir
ε
I=
R+r
Physics 150, Prof. M. Nikolic 22 Exercise: A dim flashlight As a flashlight baWery ages its emf stays approximately constant, but its internal resistance increases. A fresh baWery has an emf of 1.5V and negligible internal resistance. When the baWery needs replacing its emf is sDll 1.5V but its internal resistance has increased to 1000 V. If this old baWery is supplying 0.5 mA, what is its terminal voltage? ΔV = ε − Ir
What do we know ε = 1.5 V
r = 1000 Ω
ΔV = 1.5V − 0.5⋅10 −3 A ⋅1000Ω
i = 0.5 mA
ΔV = 1.5V − 0.5V = 1V
Physics 150, Prof. M. Nikolic 23 Resistors in series =
ΔV1
ΔV2
Req
ΔV3
The current through each resistor is the same. •  Resistors following one aMer the other •  We add them together to get the total resistance. Req = R1 + R2 + R3 + ...
•  PotenDal drops across every resistor ΔV1 = IR1 ΔV2 = IR2 ΔV3 = IR3
ΔV = ΔV1 + ΔV2 + ΔV3 = IR1 + IR2 + IR3
Physics 150, Prof. M. Nikolic 24 Resistors in parallel ΔV
=
Req
1
1 1 1
= + + +
Req R1 R2 R3
The potenDal difference (voltage) across each path is the same. ΔV1 = ΔV2 = ΔV3 = ΔV
Current is branching (like a river) è I = I1 + I2 +I3 Physics 150, Prof. M. Nikolic 25 Resistors in series and parallel (a reference set) ε = ΔV1 + ΔV2 + ΔV3
I = I1 = I 2 = I 3
R = R1 + R2 + R3
ε = ΔV1 = ΔV2 = ΔV3
I = I1 + I 2 + I 3
1 1 1 1
= + +
R R1 R2 R3
Physics 150, Prof. M. Nikolic 26 Christmas tree Old-­‐Dme Christmas tree lights had the property that, when bulb burned out, all the lights were out. How were these lights connected? In series è when one bulb burns out it breaks the circuit è no current è all bulbs are out How could you rewire them to prevent all the lights from going out when one of them burned out? Connect them in parallel connecDon è when one bulb burns out it breaks the circuit only across its own path Physics 150, Prof. M. Nikolic 27 CombinaDon circuits What to do when we have a mix of series and parallel combinaDons? 1.
Select a group of resistors connected in either series or parallel. •
In this case the two resistors on the right connected in parallel (current is branching) 2.  Calculate the equivalent resistance for the group. 3.  Go to the first step -­‐ Select a new group of resistors connected in either series or parallel in the new loop 4.  Keep going Dll you get what you want Physics 150, Prof. M. Nikolic 28 Simple analysis: all resistors are the same (a)  The two verDcal resistors are in parallel with one another, hence they can be replaced with their equivalent resistance, R/2. (b)  Now, the circuit consists of three resistors in series. The equivalent resistance of these three resistors is 2.5 R. (c)  The original circuit reduced to a single equivalent resistance. Physics 150, Prof. M. Nikolic 29 Exercise: A resistor network Three idenDcal resistors with resistance of 6 Ω are connected to a baWery with an emf of 18 V a) Find the equivalent resistance of the resistor network b) Find the current through the baWery and in each resistor 1. First, resistors R2 and R3 are connected in parallel. Their combined resistance is 1
1 1
= +
R23 R2 R3
1 1 1 1
= + =
R23 6 6 3Ω
R23 = 3Ω
2. Second, resistors R1 and R23 are now connected in series. Their combined resistance is: Req = R1 + R23 = 6 Ω + 3 Ω = 9 Ω Physics 150, Prof. M. Nikolic 30 Part (a) Part (b) Current through the baWery can be found using Ohm’s law I=
ε 18V
=
= 2A
Req 9Ω
By looking at the simplified graphs we can say that the same current flows through R1 and R23 => I1 =I23 = I =2 A This means that the voltage drop across R1 is ΔV1 = I1R1 = 2 A x 6 Ω = 12 V and across R23 is ΔV23 = I23R23 = 2 A x 3 Ω = 6 V which adds up to the 18 V provided by the baWery Since R2 and R3 are connected in parallel => ΔV2 = ΔV3 = ΔV23 =6 V => Current across R2 is I2 = ΔV2/R2 = 6 V/6Ω = 1 A => Current across R3 is I3 = ΔV3/R3 = 6 V/6Ω = 1 A Physics 150, Prof. M. Nikolic 31 AddiDonal problems (a)
(b)
(c)
Which resistors in this circuit are in parallel with each other? What is the equivalent resistance of the circuit? What are the size and direcDon of current i1 in the figure, if R1 = 9.0 ohms and R2 = 18 ohms? Physics 150, Prof. M. Nikolic 32 Circuit Analysis SoluDon (a) Resistors R
2 are in parallel Equivalent resistance 1 = 1 + 1 + 1 = 3
Re 2
R2
R2
R2
Re 2 =
R2
R2
3
(b) Resistors R 2 are in connected in series with R1
Equivalent resistance of the circuit Re = Re 2 + R1 =
Re =
(c) Current I through R
e : I =
R2
+ R1
3
18Ω
+ 9Ω = 15Ω
3
ε 12V
=
= 0.8A
Re 15Ω
Voltage drop across R 1 : ΔV1 = IR1 = 0.8A ⋅ 9Ω = 7.2V
Voltage drop across all R 2 : ΔV2 = ε − ΔV1 = 12V − 7.2V = 4.8V
Current I 1 through R 2 : I1 =
Physics 150, Prof. M. Nikolic ΔV2 4.8V
=
= 0.267A
R2
18Ω
33 Power and energy in circuits If a charge is moving through potenDal difference, the change in potenDal energy is ΔU = qΔV
A change in potenDal energy in Dme is equal to power ΔU q
P=
= ΔV = IΔV
Δt Δt
The energy/power added to the circuit by the EMF must be removed by the resistor. For an EMF source: For a resistor: Physics 150, Prof. M. Nikolic P = Iε
2
Δ
V
P = IΔV = I 2 R =
R
The resistor dissipates this energy as heat. 34 Conceptual quesDon – 40 vs 100 WaWs Q3
Which has a greater resistance, R, a 100 W or a 40 W incandescent light bulb? (The raDng of a light bulb tells you how much power it dissipates when connected to a 110 V circuit. It does not tell you how much light it emits.) A.
B.
C.
D.
A 100 W bulb has more resistance A 40 W bulb has more resistance Both the same Need more info Brightness of the bulb depends on how much power it dissipates è how much current it draws from the power grid è resistance of the bulb Physics 150, Prof. M. Nikolic 35 Q4
Conceptual quesDon – Which bulb is brighter Assume that the four light bulbs and two baWeries are idenDcal. Which circuit gives off more light? A.  The circuit with the two bulbs in parallel B.  The circuit with the two bulbs in series C.  Both the same D.  Need more info Physics 150, Prof. M. Nikolic Series: Req = 2R => Current I = ε/2R => Power dissipated P = I2R = ε2/4R Parallel: Current through every resistor I = ε/R => Power dissipated P = I2R = ε2/R 36 Q5
Conceptual quesDon – Add a wire wire Charge flows through a light bulb. Suppose a wire with negligible resistance is connected across the bulb as shown. When the wire is connected, r ε
+ A.
All the charge conDnues to flow through the bulb. B.  All the charge flows through the wire. C.  Half the charge flows through the wire, the other half conDnues through the bulb. D.  None of the above Charge/current always follows the smallest resistance path. Physics 150, Prof. M. Nikolic 37 Exercise: Power that radio A student kept his 9.0 V, 7.0 W radio turned on at full volume from 5:00 p.m. unDl 2:00 a.m. How much charge went through it? What do we know: ΔV = 9.0 V P = 7.0 W Δt from 5:00 pm to 2:00 am Δq
I=
Δt
Δq = IΔt
Time: t = 9 h = 9 × 3600 s = 32400 s
We can find the current from power dissipated by the radio P = IΔV
I=
P
ΔV
I=
7W
= 0.78A
9V
Δq = IΔt = 0.78A ⋅ 32400s = 25272C = 2.53⋅10 4 C
Physics 150, Prof. M. Nikolic 38 Kirchhoff’s rules – JuncDon rule The sum of the current entering any juncDon must be equal to the sum of the currents leaving that juncDon ∑I
in
= ∑ I out
A juncDon is a place where two or more wires (or other components) meet. Physics 150, Prof. M. Nikolic 39 Kirchhoff’s rules – loop rule The algebraic sum of the potenDal difference in any loop must equal zero. ∑ ΔV = 0
For any circuit we must calculate this rule for the minimum number of loops such that each circuit element is crossed at least once. The direcDon of the loop is your choice. Physics 150, Prof. M. Nikolic 40 Kirchhoff’s rules – loop rule Sign convenDons Physics 150, Prof. M. Nikolic 41 Exercise: Circuit analysis I Calculate the current through each ideal baWery in the figure. I
a
R1
I
R2
I1
+
ε1 −
I
I2
I +
R1
−
I
b
R2 = 2Ω
ε1 = 2.0V
ε 2 = ε3 = 3.0V
+
−
ε3
II R1
ε2
R1 = 1Ω
Physics 150, Prof. M. Nikolic R1
1. Since the direcDons of the currents are not given – assume it yourself I2
JuncDon rule: I = I1 + I 2
2. Draw two loops and choose the direcDons of both Loop I: ε1 − IR1 − I1R2 − ε 2 − IR1 = 0
I2
2 − I − 2I1 − 3− I = 0
Loop II: ε 2 + I1R2 − I 2 R1 − ε3 − I 2 R1 = 0
3+ 2I1 − I 2 − 3− I 2 = 0
42 Exercise: Circuit analysis I I = I1 + I 2
2 − I − 2I1 − 3− I = 0
2I1 = 2I 2
3+ 2I1 − I 2 − 3− I 2 = 0
I1 = I 2
SubsDtute this into the first equaDon I = I1 + I 2 = 2I1
And place this into the second equaDon 2 − 2I1 − 2I1 − 3− 2I1 = 0
6I1 = −1
I1 = −0.167A
I 2 = I1 = −0.167A
I = 2I1 = −0.333A
Physics 150, Prof. M. Nikolic Current is a scalar – it has to be a posiDve number. Minus signs in the current values mean that all the currents actually flow in the direcDons opposite to the ones we assumed. 43 Exercise: Circuit analysis II a)  Apply the juncDon rule at each juncDon and write the equaDons that result. b)  Choose two loops and write the equaDons resulDng from the loop rule. c)  Find i, r and E Loop I Loop II Loop I: JuncDon rule: ∑I
in
= ∑ I out
i = 1A + 2A = 3A
Physics 150, Prof. M. Nikolic Loop II: ε +1A ⋅1Ω − 2A ⋅ 3Ω = 0
ε = −1V + 6V = 5V
12V − ir −1A ⋅1Ω − ε = 0
12V − 3A ⋅ r −1V − 5V = 0
6V = 3A ⋅ r
r = 2Ω
44 What if capacitors occur mulDple Dmes in a circuit? Capacitors in parallel ε
C1 C2 C3 C4 •  The potenDal difference (voltage) across capacitors in parallel must by the same •  The charges on the plates add up ΔV1 = ΔV2 = ΔV3 = ε
n
Qeq = Q1 + Q2 +...
Ceq = C1 + C2 + … + Cn = ∑ Ci .
i =1
ε Ceq = ε C1 + ε C2 +...
Physics 150, Prof. M. Nikolic 45 Capacitors in series C1 C2 C3 C4 +Q -­‐Q +Q -­‐Q +Q -­‐Q +Q -­‐Q ε
•  For capacitors in series the charge on the plates is the same. ε = ΔV1 + ΔV2 + ΔV3 + ...
Q Q Q
= + +..
Ceq C1 C2
Physics 150, Prof. M. Nikolic n
1
1 1
1
1
= + +…+
=∑ .
Ceq C1 C2
Cn i=1 Ci
46 Exercise: Equivalent capacitance Find the equivalent capacitance of the following circuit: 4.00 nF 6.00 nF 3.00 nF 2.00 nF C1 = 2nF + 3nF = 5nF
12.0 nF 8.00 nF 3.00 nF 3. 4.0 nF, 6.0 nF and 3.0 nF capacitors are connected in series 1
1
1
1
9
=
+
+
=
C3 4nF 6nF 3nF 12nF
C3 =
1. 2.0 nF and 3.0 nF capacitors are connected in parallel 2. These two are in series with 8.0 nF capacitor 1
1
1
13
=
+
=
C2 8nF 5nF 40nF
C2 =
40
nF = 3.08nF
13
12
nF = 1.33nF
9
Physics 150, Prof. M. Nikolic 47 Exercise: Equivalent capacitance Find the equivalent capacitance of the following circuit: 1.33 nF 4. C2 = 3.08 nF and C3 = 1.33 nF capacitors are connected in parallel C4 = 3.08nF +1.33nF = 4.41nF
12.0 nF 3.08 nF 5. This is in series with 12.0 nF capacitor 1
1
1
1
=
+
= 0.31
Ceq 4.41nF 12nF
nF
Ceq = 3.22nF
Physics 150, Prof. M. Nikolic 48 RC circuits Suppose we put a capacitor and a resistor in series with a baWery When we close the switch Switch •
R •
-­‐ + + Current will flow Charge will start to build up on the capacitor C -­‐ ε As more charge builds up on the capacitor, it gets harder for current to flow (repelled by the charges already on the capacitor plate) •
stored charge goes up •
amount of current flow goes down Physics 150, Prof. M. Nikolic 49 Charging a capacitor (
VC (t ) = ε 1− e
(
−t
τ
Q C (t ) = Cε 1− e
Physics 150, Prof. M. Nikolic )
I (t ) = I 0 e
−t
ε
I0 =
and τ = RC
R
τ
)
−t
τ
τ=RC is the Dme constant 50 Exercise: Charging a capacitor An iniDally uncharged 70-­‐µF capacitor is hooked up to a 560-­‐kΩ resistor. What is the charge on the capacitor 8 s aMer making the connecDon with an ideal 9.00-­‐V baWery? (
Q C (t ) = Cε 1− e
−t
τ
)
−8s
$
'
3
−6
Q C (t ) = 70 ⋅10 F ⋅ 9V &1− e 560⋅10 Ω⋅70⋅10 F )
%
(
−6
−6
(
Q C (t ) = 630 ⋅10 C 1− e
−8s
39.2s
)
Q C (t ) = 630 ⋅10 −6 C ⋅ 0.18 = 116.310 −6 C
Physics 150, Prof. M. Nikolic 51 Exercise: Charging a capacitor A capacitor, C = 3.1 µF, and a resistor, R = 5.3 kΩ, are connected to a baWery, ε = 15 V, as shown. At t = 0 the switch is closed. At what Dme is the current across the resistor equal to 25% of its maximum original value? R What do we know: C = 3.1 µF R = 5.3 kΩ ε = 15 V 0.25I 0 = I 0 e
−t
I (t ) = I 0 e
−t
τ
When current is at 25% of its original value => I = 0.25I0 τ
0.25 = e
−t
S τ
Now we have to deal with a logarithmic funcDon, but t
= − ln 0.25
τ
C ε
A = e−B ⇒ B = − ln A
Natural logarithm – built in your calculator Time constant τ = RC t = −τ ln 0.25
Physics 150, Prof. M. Nikolic t = −3.1⋅10 −6 F ⋅ 5.3⋅10 3 Ω⋅ ln 0.25 = 22.78⋅10 −352 s
Measuring voltage and current •  Ideal voltmeter draws no current – connected in parallel (infinite resistance) •  Ideal ammeter causes no change in potenDal change – connected in series (zero resistance) V
A
Physics 150, Prof. M. Nikolic 53 Review Electric current Resistors in series Δq
I=
Δt
Req = R1 + R2 + R3 + ...
The current in the wire is: Δq eN e
I=
=
= neAvd
Δt
Δt
Resistance Temperature dependence L
R=ρ
A
ρ = ρ0 (1 + α (T − T0 ))
Ohm’s law R = R0 (1+ α (T − T0 ))
ΔV = ΔV1 + ΔV2 + ΔV3 = IR1 + IR2 + IR3
Resistors in parallel 1
1 1 1
= + + + ...
Req R1 R2 R3
ΔV1 = ΔV2 = ΔV3 = ΔV
Current is branching (like a river) è I = I1 + I2 +I3 ΔV = IR
Physics 150, Prof. M. Nikolic 54 Power for an EMF P = Iε
Review Kirchhoff’s rules For a resistor: ΔV
P = IΔV = I R =
R
2
JuncDon rule ∑ I in = ∑ I out
2
Loop rule Capacitors in series Charging a capacitor n
1
1 1
1
1
= + +…+
=∑ .
Ceq C1 C2
Cn i=1 Ci
Capacitors in parallel n
Ceq = C1 + C2 + … + Cn = ∑ Ci .
i =1
Physics 150, Prof. M. Nikolic ∑ ΔV = 0
(
VC (t ) = ε 1− e
I (t ) = I 0 e
I0 =
−t
−t
τ
)
(
Q C (t ) = Cε 1− e
−t
τ
)
τ
ε
and τ = RC
R
55