# R C V + - In the circuit at left we have a battery with voltage V, a resistor with resistance R and a capacitor with capacitance C. Which direction should.

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R C V + - In the circuit at left we have a battery with voltage V, a resistor with resistance R and a capacitor with capacitance C. Which direction should the arrow indicating conventional current be drawn and which direction do the electrons in the current flow? The two possibilities are labeled 1 and 2 on the diagram. a)Current and electrons in direction 1 b)Current and electrons in direction 2 c)Current in direction 1, electrons in direction 2 d)Current in direction 2, electrons in direction1 1 2

Correct Answer – D For historical reasons we always point electric field lines in the direction from the positive charge to the negative charge. Positive charges would move away from a positive charge towards a negative charge, so they would move from the positively charged pole of the battery (the long bar on the battery symbol) to the negatively charged pole (the short bar). Therefore to be consistent with our field line convention we show the current arrow on a circuit diagram pointing in the direction the positive charges would flow. This is called the direction of “conventional” current. Of course the real current consists of electrons, which are negative and therefore move from the negative pole to the positive pole, the opposite direction to conventional current. Note that in practice you cannot tell the different between a positive current flowing counterclockwise and a negative current flowing clockwise.

R C V + - Let’s introduce a switch into the circuit. When we press the switch the circuit will be closed. Before we close the circuit, no current can flow, because there is no conducting route between the poles of the battery. Now, when we first close the switch and there is no charge at all on the capacitor, how much current, I o do we expect to start flowing? a)I o = Q/V b)I o = V/R c)I o = C R d)I o = V – C/R IoIo

Correct Answer – B If there is no charge on the capacitor then there cannot be any voltage across it either, since V = Q/C for a capacitor. Therefore the whole voltage of the battery must be equal to the voltage across the resistory (which is the only other circuit element) by the “loop rule.” Now if the voltage across the resistor R is V, then Ohm’s Law tells us that the current through the resistor is I o =V/R. If the current flowing through the resistor is I o, and since there are no junctions in the circuit (nowhere else for the current to go) the junction rule tells us that I o is the current through every part of the circuit.

R C V + - I As the current flows in the circuit, it will obviously tend to “charge up” the capacitor, by depositing charge on its plate. After a while let’s suppose an amount of charge Q has been deposited on the plates. How much current, I do we expect to be flowing in the circuit now? Let’s call the current that flowed earlier on, immediately after the switch was closed I o. a)I = I o b)I < I o c)I > I o d)I = 0 +Q -Q

Correct Answer – B Since the negative pole of the battery is sending out negative electrons, it is negative charge that piles up on the side of the capacitor closest to the battery’s negative pole. Since it is the positive pole of the battery which is sucking in electrons, the capacitor plate closest to it is losing electrons and becoming positively charged. Note that no current is actually crossing between the two plates of the capacitor. The result is that after a while a voltage builds up on the capacitor, V c = Q/C. But note that this voltage acts to oppose the voltage of the battery. We have positive connected to positive and negative connected to negative, which everyone who has ever plugged batteries into a toy on Christmas morning knows is a no-no. So the voltage across the battery should be less than it was before, and by Ohm’s law the currect flowing through it will be less also.

R C V + - I +Q -Q What is the current flowing through the battery when the situation is as shown, with a charge Q on the plates of the capacitor? a)I = V/R – Q b)I = VQ/R c)I = V/R – Q/(RC) d)I = V/R - QR

Correct Answer – C The voltage across the capacitor is V c = Q/C, and this voltage has the opposite sign to the battery voltage. The loop rule tells the total voltage around the circuit must be zero, so V - V c + V R = 0, where V R is is the voltage across the resistory. Therefore V R = V - V c. (I’ve adjusted the sign of V R to make it positive, since it really doesn’t matter). Now the current through the resistor (and therefore throughout the circuit) is I= V R /R by Ohm’s Law, therefore I = V/R - V c /R = V/R – Q/(RC). Note that V/R = I o is simply the initial current after the switch was closed, so this equation reads I = I o – Q/RC. Therefore as the charge Q on the capacitor plates increases, the current flowing in the circuit decreases.

Now although our equation I = I o – Q/(RC) tells us how the current decreases as the charge on the capacitor plates increases, it doesn’t tell us how quickly all this happens. We’d like to know something about how long this all takes, but first we need to learn more about our equation. The only thing that changes to make I change, is Q, which is divided by the quantity RC. What does this RC quantity mean? The first question we need to ask is, what are its units? What are the units of RC? a)Amps b)Newtons c)Volts d)Seconds e)Meters

Correct Answer – D So, oddly enough, resistance times capacitance is a measure of time! There are two ways of figuring this out. The first is to look at our equation I = I o – Q/RC. I is measured in Amps, Q in Coulombs, and the definition of Amps is Coulomb/seconds. Therefore for our equation to mean anything (if it is not to compare apples to oranges) then RC must have units of seconds. You can check this as follows. The units of R C are Ohms x Farads. Since V=I R we know that R = V/I so Ohms are simply Volts/Amps. We also know that C=Q/V so Farads are Coulombs/Volts. Therefore Ohms x Farads = (Volts/Amps) x (Coulombs/Volts) = Coulombs/Amps = Coulombs/(Coulombs/seconds) = seconds.

R C V + - I +Q -Q RC is called the “time constant” of the circuit. We usually label it  =RC. It is roughly the amount of time it takes for the capacitor to charge up. If we would like the capacitor to charge up quickly, what must we do? a)Keep R and C small b)Keep R and C large c)Keep R small but C large d)Keep R large but C small

Correct answer – A Not very surprisingly small C speeds up the charging process, because small C means less charge Q per unit voltage V on the plates, and it surely takes less time to build up a smaller total charge. Small R also helps, because the smaller R is the larger the initial current is, and therefore the more charge that is flowing to the plates.

R C V + - I +Q -Q The final question is whether the charge is built up evenly over time, or whether the build up is slow at first and then picks up, or the opposite. Given what we know about the current, which of the following sounds right? a)The charge builds up slowly at first and then more and more quickly b)The charge builds up evenly over time c)The charge builds up quickly at first and then more and more slowly

Correct answer – C Because the current is large initially, it is possibly to build up a lot of charge quickly at the beginning. As the charge builds up, it opposes the current, which decreases, and so less and less charge is reaching the plates after a time equal to 1 or 2 time constants  = RC. The maximum amount of charge, which the capacitor approaches but never quite reaches, is given by our familiar equation Q = CV. Q t 

I t  V/R The reason the charge never actually reaches the maximum possible value is that the larger the charge on the capacitor gets, the smaller the current bringing the charge to the capacitor left. This follows a similar kind of exponential curve to the charge curve, except that it starts off large, falls quickly and then slows down to approach zero, never quite reaching it. The rate of fall of is governed by the same time constant  = RC.

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