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Today: - RC circuits (time-varying currents) - Electric Eel RC Circuits Thus far, we dealt only with circuits in which the currents did not vary with time. Here we begin a discussion of time-varying currents. Charging a capacitor: We are now going to explore the time dependence of the charge on a capacitor. If we start with a circuit with a resistor and a capacitor in series with a emf, but with the circuit broken by an open switch, the capacitor is uncharged, since there is no potential difference across it.

At some time, t0, we close the switch, let us now examine, using the loop rule:

V i ( t ) R

q( t ) 0 C

But this equation relates two quantities that vary over time: i(t) and q(t). But, since the charge on the capacitor, q(t), comes from the current, i(t)=dq/dt, we can relate these two as well: q( t ) q (t ) R 0 dt C q( t ) 1 q( t ) 0 dt RC R

This is a differential equation, you do not need to know how to solve. But you need to be able to apply the solution. The solution to this equation is

PES 1120 Spring 2014, Spendier

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t t q(t ) C 1 e RC Q0 1 e RC

Let’s also check our boundary values of q(t) at t=0 and as t = ∞: 0 RC q(0) C 1 e C 1 1 0 q( ) C 1 e RC C 1 0 C Q0

Then we can take the derivative to solve for the current: dq(t ) RCt i(t ) e dt R And do the same for the current: 0 i(0) e RC 1 R R R i( ) e RC 0 0 R R

Note how the capacitor acts at these extremes of times. At t=0, immediately after the switch is closed, current flows like the capacitor is just a wire. At very long times after the switch is closed, the capacitor acts like a broken circuit, so no current passes through that part of the circuit.

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Using q = CV, we find that the potential difference VC(t) across the capacitor during the charging process is t q RC Vc (t ) 1 e C at t = 0: VC(t) = 0 at t = ∞: VC(t) = ε

Time Constant The factor RC in the denominator of the exponential is commonly called the RC time constant, and determines the rate at which the capacitor charges (and discharges, as we will soon see). τ = RC t q(t ) Q0 1 e

Example 1 A 10 M resistor is connected in series with a 1.0 μF capacitor and a battery with emf 12.0 V. Before the switch is closed at time t = 0, the capacitor is uncharged. a) What is the time constant? b) What fraction of the final charge Q0 is on the capacitor at t = 46 s? c) What fraction of the initial current I0 is still flowing at t = 46s?

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Discharging a capacitor Now, let’s start with a fully charged capacitor in an open circuit with no battery. Initially, there is charge q0 on the capacitor. For t < 0, the switch is open: - the potential difference across the capacitor is given by Vc = q/C. - the potential difference across the resistor is zero because there is no current flow, I=0. At some time, t=0, we will close the switch. The capacitor will begin to discharge and charge will begin to flow around the circuit.

The charged capacitor is now acting like a voltage source to drive current around the circuit. When the capacitor discharges (electrons flow from the negative plate through the wire to the positive plate), the voltage across the capacitor decreases. The capacitor is losing strength as a voltage source. Applying the loop rule by traversing the loop clockwise, the equation that describes the discharging process is given by

V

q( t ) i (t ) R 0 C

The current that flows away from the positive plate is proportional to the charge on the plate,

I

dq dt

The negative sign in the equation is due to the fact that the charge on the positive plate is decreasing as more positive charges leave the positive plate. Thus, charge satisfies a first order differential equation:

dq(t ) 1 q( t ) 0 dt RC

and the solution to this equation

q(t ) Q0 e

t RC

Q0 e

t

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To find the time dependence of the current, we just have to take the time derivative of this solution:

i(t )

Q t q(t ) 0 e dt RC

The current must be negative since the positive charge moves in the opposite direction when a capacitor is discharged then when it is charged.

Example 2: The resistor and capacitor of example 1 are reconnected. The capacitor has an initial charge of 5.0 μC and is discharged by closing the switch at t = 0. a) At what time will the charge be equal to 0.5 μC? b) What is the current at this time?

PES 1120 Spring 2014, Spendier

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Electric Eel http://www.youtube.com/watch?v=967BH1Y4v-s (old) The electric eel (Electrophorus electricus), lives in rivers of South America. It lives on fish which the eel kills by electric shocks. The electric eel generates the voltage in special sets of cells called electroplaques. How do electric eels stun or kill fish without stunning or killing itself? The eels generate current with electroplaques, biological cells that are essentially batteries, with an emf source and resistance. Each eel has approximately 140 rows of these cells that run the length of their body. Each row contains nearly 5000 electroplaques. Each electroplaque has an emf of 0.15V and a resistance of 0.25Ω. The resistance of the water (and fish) between the eel’s head and tail to be 800Ω. There is a current that would pass through the prey (ouch) and a current that passes through the eel. Essentially they form a complete circuit- the current leaves the eel, passes through the water and fish and then back to the eel. But, if we simply take the eel as a single pathway, then the current would be the same through the predator and prey, and that’s no good. The key is that the eel has many rows over which the current can pass. So, instead of a single path or branch, there are actually 140. This means that the current divides among these rows, thereby reducing the current that goes through any given part of the eel. (Also, I’m going to guess that much of the current passes over the surface of the eel, rather than through its body. This would also reduce the risk of self-inflicted injury.) We will treat the eel and fish system as an idealized circuit. In each row, there are 5000 electroplaques. Each one has a resistance and emf. Since the electroplaques in a single row are in series we can replace the 10,000 “components” with simply two- an equivalent emf source and resistance. The emf for each row is going to be Vrow= 5000* Velectroplaque= 5000* 0.15V= 750V For the resistance; Rrow= 5000* Relectroplaque= 5000* 0.25Ω= 1250Ω Once we recognize that each of the 140 rows can be replaced with an equivalent emf source and resistor, we end up with something that looks like this:

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(Okay, so I got lazy and didn’t draw all 140 rows; just use your imagination.) This is now bordering on the doable. We still have lots of components and branches, but we can employ Kirchoff’s rules to find the current in each row as well as the current through the prey. Let’s redraw the circuit, labeling the currents.

Notice that the currents in each row are equivalent. (Even if you do not see that at this stage, and instead work with a different current for each row, I1, I2, I3, etc., you’ll get the same result in the end.) But,… the current through the fish is different. Now we’ll be able to write down two equations and solve for the two unknowns- current through the fish and the current through each row along the eel. Current/ junction rule: 140 * I= Ifish Next we need to select a loop and direction of travel:

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I’ve picked one of the rows and the fish. Potential/ loop rule: -1250Ω I + 750V-800Ω Ifish=0 With these two equations we can now solve for the currents. Ifish ≈ 0.9A and I≈ 7 x 10-3A Does this make any sense? In an earlier lecture I mentioned that 100 mA is a lethal shock for humans. So, our current through the fish does make sense. Also, the current through the eel, is much lower. (It might cause some pain to the eel, but I’m sure that is more appealing than starving to death.) This is the beauty of having many separate rows through which the current can travel.