This circuit is equivalent to the previous one
6V
2Ω
Oh no!
Short circuit
A: 6 A
B: 3 A C: 2 A
E: Ammeter will "fry"
D: 0 A
With R_equiv=15 Ohm, it means 30 V / 15 Ohm = 2 Amps flows out of the baAery. When that reaches the “juncHon” at the top, it splits. Since the two 10 Ohm parallel resisters are equal, the current splits evenly, so 1 A goes through the Ammeter. Each of the 3 resistors is 10 Ohms.
What does the Ammeter “A” read?
A
Requiv=15 Ohm
A: 30 Amps
B: 3 A
E: Something else!
30 V
C: 2 A
D: 1 A
Ammeter has a finite r in series with the 10 Ohm to the leN of it. This INCREASES the r of that top line, and in turn increases the effecHve R_total (by a small bit). So less current comes out of the baAery. Furthermore, when reaching the juncture at the top, current doesn’t split perfectly evenly: slightly LESS current will flow through the slightly LARGER resistance of the top line. All together, the Ammeter gets a slightly smaller “share” of a slightly smaller total, so overall, slightly LESS than ideal! If the ammeter “A” is real (not ideal), how
does its reading compare to our “ideal”
answer in the previous question?
A
30 V
A: “Real” reading is slightly LESS than ideal
B: “Real” reading is slightly MORE than ideal
C: Makes no difference at all!
CT 29.30
Back to ideal meters! Each of the 3
resistors is 10 Ohms.
What does the Voltmeter read?
See next slides for explanaHon. OR: Use our previous result from above, that I through the Ammeter is 1 Amp, so Delta V across the resistor next to the Ammeter is 1 Amp * 10 Ohms = 10 Volts, and that’s what the voltmeter is “across”! V
A
30 V
Recall, Requiv=15 Ohm
A: 30 V
B: 20 V
E: Something else!
C: 10 V
D: 1 V
CT 29.30
V
A
IbaAery? Recall, Requiv=15 Ohm
IbaAery = 30 V/ 15 Ohm = 2 A 30 V
CT 29.30
V
0V
A
0V
0V
2 A 0V
30 V baAery CT 29.30
V
0V
A
0V
0V
2 A V(here)=? 0V
30 V baAery CT 29.30
V
?? V
0V
A
0V
0V
2 A 30 V 0V
CT 29.30
V(here) = 30V – ΔV = 30 V -­‐ (2A*10 Ω) = 30 V -­‐ 20 V = 10V V
0V
A
0V
0V
2 A 30 V 0V
See next slide: If we consider the Voltage change starting
at Point P and going all the way around the
circuit loop back to Point P, what is the grand
total ΔV all the way around?
Α.
Β.
C.
D.
E.
ΔV = +6.0 Volts
ΔV = +12 Volts
ΔV = +18 Volts
ΔV = 0.0 Volts
Something different!!
Ibat
RA
IA
6V
Point P
RB
IB
Kirchhoff’s Voltage law If we consider the Voltage change starting
at Point P and going all the way around the
circuit loop back to Point P,
the grand total ΔV all the way around is zero!
ΔV (all the way around)
= Sum (individual ΔV’s)
= 0.0 Volts
Ibat
V
RA
IA
6V
Point P
RB
IB
Which is the correct current equation for this
circuit? (Kirchoff’s current rule)
B: Current in = I1 + I3, Current out = I2. I1
I2
I3
A) I1=I2+I3 B) I2=I1+I3
C) I3 = I1+I2
D) MORE than one of these is ok !
See next slides! (The answer is B, neither one changes at all! This is quite counterintuiHve, but look at the slides to follow and convince yourself. A stays off whether that switch is open or not, and B stays on whether that switch is open or not! Some people really want switches to turn bulbs on or off ;-­‐) But, it depends on the circuit! Imagine for a second that B was not there. This would be like having two baAeries side by side, with a bulb between their + poles. No current flows, the light bulb is dark! What happens to the
(identical) bulbs when
the switch is closed?
A: A stays same,
B changes
B: A stays same,
B stays same
C: A changes, B changes D: A changes, B stays same VHERE=? 0 V VHERE=? 0 V 0 V VHERE=? 12 V
12 V
12 V
12 V
What happens to the
(identical) bulbs when the
switch is closed?
12 V
12 V
12 V
12 V
What about now?
What about now?
What about now?
A
VHERE=? 12 V
B
In this situaHon, both bulbs are glowing, equal to each other, but ¼ the brightness of what bulb B was doing before! (Before it had 12V across it, now it only has 6 V across it, and power goes like Delta V^2/
R, so ½ the Delta V means ¼ the brightness. What about now?
A
6 V (If A and B are idenHcal) 12 V
B
-­‐A is brightest, it has the FULL voltage across it. -­‐D is less bright than A (because it doesn’t have full baAery voltage across it, E and F (for isntance) must have some “share” of the total baAery voltage across them, and the SUM of voltage drops across D + E + F must add up to the voltage of the baAery!) -­‐ E and C are in parallel, so equally bright. -­‐ But they are each LESS bright than D, because all the current that goes through D has split up, only PART goes through E, and less current means less bright by P = I^2 R. -­‐ F is clearly MORE bright than E or C, because all the current in C AND E go together to go through F. (More current, brighter!) -­‐ F is LESS bright than D. That took a sec for me to see, but note that SOME of the current that went through D heads over to B, leaving a bit less to go through C+E and thence through F. -­‐  B is brighter than F (because the voltage drop across it is the voltages across E AND F combined!) Puqng it all together: A > D > B > F > C=E. Wow! CT 29.22
All the bulbs are identical.
Which is brightest?
Going FROM a TO b That’s going from “posiHve side of baAery to negaHve side” , which is going DOWN in voltage, thus Delta V is negaHve. Going FROM a TO B That’s going “against the flow of current”. But current always goes “downhill” in voltage, so we must be going Uphill, and thus Delta V is POSITIVE (IR.) So the answer is B As you walk past the two circuit elements
shown, moving FROM a TO b, what is
ΔV = V(b)-V(a) in each situation?
Situation
#1
#2
A) +V, +I R
B) -V,
+IR
C) +V, -IR
D) -V,
-IR
Look at the black dot at the top: I1 and I2 are exiHng, I3 goes around the corner, through V1, and IN to the dot. So I3 enters, I1 + I2 exits, the answer is C. The current directions have been guessed as
shown. Which is the correct current equation
I1
for this circuit?
R1
V1
R2
I2 Loop 1.
R3
A)  I2=I1+I3
I3
B) I1 = I2+I3 C) I3 = I1+I2
V2
B=C, they are in series E=F, they are in parallel D > E=F, because all the current that goes through D gets split, E and F each get HALF the current. Now the hard part: in the second circuit, that E+F parallel combo has resulted in LESS effecHve resistance than C alone, so R_eff is smaller in the second circuit, so MORE current flows out of the second circuit’s baAery. It all goes through D, so D must be brighter than B=C. Finally, if D is brighter than B, that means there is more delta V across it. But the TOTAL Delta V is “V” in both circuits, so if there is more “drop” across D than B, there must be LESS “drop” across the E+F combo than across C. So E+F are dimmer than C. BoAom line: D > B=C > E=F, answer is C All bulbs are idenHcal…rank the brightness of the bulbs. A)
B)
C)
D)
E)
B=C > E=F > D B=C > D > E=F D>B=C>E=F E=F > B=C > D Something else! See next slides for a more pictorial explanaHon: All bulbs are idenHcal…rank the brightness of the bulbs. B=C D> E=F Resistance of the the second circuit is LESS So, more current in the second circuit So, D>B=C. All bulbs are idenHcal… V More or less Than V/2?? V/2 See earlier, Delta V across bulb D is MORE than across B, so the The actual value of voltage at the locaHon of the dot in the right hand picture is LESS than V/2 0 0 0 Recall, D is brighter than B! 0 CT 29.22
Adding one more R in parallel REDUCES the effecHve resistance, And smaller R_eff means MORE current flows out of the baAery. It all goes through A, so A is BRIGHTER in the right hand circuit than in the leN hand circuit! Suppose you add bulb “D” to the circuit on the leN, what happens to the brightness of bulb A? A) Increases B) Decreases C) Stays the same CT 29.22
All the bulbs in the circuit below are idenHcal. Rank the brightness of the bulbs… A has all the current, so it must be brightest. Current “splits” at the juncHon, but we have learned that when you reach a parallel juncture, MORE current heads down the path with LESS resistance. That would be the “B” path. So I get A > B > C=D. A)
B)
C)
D)
A>B=C=D B>C=D>A A>B>C>D A>B>C=D E) Other! CT 29.22
All bulbs are idenHcal, what happens to the brightness of Bulb C when you close the switch? BEFORE: A is off. B and C are in series, so R_eff = R+R = 2R AFTER: A and B are now in Parallel, making R_eff(A+B) = R/2, and these are in series with C, making R_eff(all three) = R/2 + R. Thus, R_eff went DOWN (from 2R to 1.5 R) and thus total current through the baAery went UP. But it all goes through C no maAer what! So, C got BRIGHTER when you closed the switch. (And, A turned on! And, B got DIMMER) A
12 V
A) Increases B) Decreases C) Stays the same switch
B
C
CT 29.22
All bulbs are idenHcal, what happens to the brightness of Bulb A when you close the switch? See above, it increased, from nothing to something! A
12 V
A) Increases B) Decreases C) Stays the same switch
B
C
CT 29.22
All bulbs are idenHcal, what happens to the brightness of Bulb A when you close the switch? Closing the switch “short circuits” bulb C, which REDUCES the effecHve resistance of the C+D combo, which REDUCES the overall resistance. Less total R means MORE current through A. So it got brighter! A) Increases B) Decreases C) Stays the same Regarding B: We decided A got brighter. Which means Delta V across A is more. But, since 120 V = Delta V across A + Delta V across B, then the voltage across B got corespondingly smaller. So B got DIMMER. Regarding bulb D: Before we flipped the switch, there was MORE overall total resistance, and so LESS current going through A. Furthermore, at the juncture, A SMALLER fracHon will go through the C+D combo, because current favors the lower resistance side of parallel juncHons. So Bulb D was geqng “less fracHon of less current” before. Thus, aNer flipping the switch down, D got Brighter. So the answer must be C! CT 29.22
All bulbs are idenHcal, what happens to the brightness of Bulbs B and D when you close the switch? A) Bulb B gets brighter, Bulb D stays the same B) Bulb B stays the same, Bulb D gets brighter C) Bulb B gets dimmer, Bulb D gets brighter D) Both stay the same brightness E) Something else! CT 29.22
What happens to the brightness of Bulbs B and D now when you close the switch? This circuit is different than the previous one, there is no “Bulb A” anymore. That means that B is ALWAYS seeing the full 120V, no maAer what the switch does. So B stays the same. However, flipping the switch means C is “short circuited out”, and effecHvely D goes from having 60 V across it to having the full 120 V across it, so D got brighter. The answer is now B. A) Bulb B gets brighter, Bulb D stays the same B) Bulb B stays the same, Bulb D gets brighter C) Bulb B gets dimmer, Bulb D gets brighter D) Both stay the same brightness E) Something else! What is the correct voltage equation for loop 1?
I1
We aren’t fussing with this this semester, but going around the loop #1, we start by R2
going DOWNstream in voltage across R2 (so V1
–I2R2), then we go DOWN in voltage as we pass baAery V2 going from + side to – side, R3
so that’s –V2, then we go UP in voltage as we pass R1 (because we are going Upstream in voltage), this is +I1 R1. I3
And the sum of all changes is zero. A)  -I2R2-V2-I1R1=0
Looks like answer B is correct. C)+I2R2+V2-I1R1=0
(Oh, but answer C is the same equaHon mulHplied through by negaHve one, so technically answer C is ok too!) R1
I2 Loop 1.
V2
B) -I2R2-V2+I1R1=0
D) +I2R2+V2+I1R1=0