Physical Sciences 3: Assignments for Feb Homework #2: Capacitance and DC Circuits

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1 Physical Sciences 3 Feb 7 1, 13 Physical Sciences 3: Assignments for Feb 7 1 Homework : Capacitance and DC Circuits 1 Voltage divider (1 pts) You have an ideal battery with E = 1 V and a resistor with R = 1 kω, along with some wire You would like to connect the resistor to the battery such that the voltage across R is 5 V This will reuire adding a second resistor to the circuit Draw a circuit diagram indicating how you would connect the battery and two resistors to achieve the desired result What must be the resistance of the extra resistor? We can t connect the two resistors in parallel with the battery, because then the voltage across each would be 5 V, and we want the voltage across the 1-kΩ resistor to be 5 V So we have to connect them in series, as shown at left The voltage across the first resistor R is V a V b, and the voltage across the extra resistor R x is V b V c We want the voltage across R to be 5 V, so the current through R must be I = ΔV R = 5 V 1 Ω = 5 ma from Ohm s law This same current must pass through the other resistor because they re in series with one another We also know that two resistors in series act like a single resistance with R euiv = R + R x, so the current must be: I = 1 V R + R x Setting that eual to 5 ma and solving gives R + R x = 4 Ω, so the extra resistance must be R x = 14 Ω 1

2 Physical Sciences 3 Feb 7 1, 13 Membrane model (4 pts) In class we discussed a circuit that can model some of the electrical characteristics of a cell membrane Let s analyze this circuit (at right) a) Consider the case in which the capacitor starts out uncharged, and then both switches S 1 and S are turned on that is, you turn on both the ion pumps and the ion channels Write the Kirchhoff s Law euations for this circuit (Please provide a clear circuit diagram with your currents and loops labeled) Junction A: I 1 = I + I 3 Big loop: ε I 1 = Left loop: ε I 1 C = Right loop: C = b) Use your euations from part (a) to derive the following differential euation for the charge on the capacitor: dt = E in which we define an euivalent capacitance! C C = C The charge on the capacitor is related to current I 3 above: I 3 = dt = I 1 Solving for the currents: I 1 = ε C But, we ve defined C! = I = C I 3 = dt = I 1 = ε C ( = ε C C ( = ε C C Plugging this in: dt = ε C ( = ε C ) (

3 Physical Sciences 3 Feb 7 1, 13 c) Show explicitly that this differential euation has a solution: (t) = 1 e t C ( ), by plugging that solution into the euation and showing that it works Then find an expression for First we need to take the derivative of with respect to t: dt = d 1 e t/ ( C ) dt 1 ( ) = Plugging this in to our statement from part b): dt = ε C C ( e t/ C 1 e t/ C! =? t/r C (! 1 ) ε 1 e max =? ε C = ε! C C +! max e t/r1 The euation (t) = 1 e t C = ε C ( ) is a solution if we define = ε d) We can think of the value of as modeling the speed of the pumps: if the pumps are very fast, then will be small Likewise, the value of models the speed of the ion channels: if the channels allow ions to flow uickly, then will be small In a neuron, the ion channels are uite fast, while the pumps are relatively slow What does this imply for the relative magnitudes of and? What will be the approximate limiting charge in this case if both the pumps and the ion channels are active? What about when the ion channels are turned off ( ), and only the pumps are active? In the neuron the ion channels are fast while the pumps are slow In our model, this would mean that >> The limiting charge in this case would be: If, then the limiting charge would be: R = ε C = ε C ( ε C ( C R = ε C = ε C ( = ε C ( R ( εc 1 +1( 3

4 Physical Sciences 3 Feb 7 1, 13 3 Energy in a discharging capacitor (3 pts) Consider a capacitor C with an initial charge At time t =, the switch S is closed and the capacitor begins to discharge through resistor R a) How does the total energy dissipated in the resistor relate to the total energy initially stored in the capacitor? Explain The system (ie, the circuit) only contains the resistor and the capacitor With energy conservation in mind, all of the initial stored electrical potential energy must be converted to heat through ohmic heating in the resistor b) Derive an expression for the energy stored in the capacitor as a function of time Express your answer in terms of, C, R, and t The energy stored in the capacitor can be expressed as: U = 1 C For a discharging capacitor, we know that the charge as a function of time can be written Q t so we can just plug that into the first expression to get U(t): 4 Q ( ) = e t U ( t) = 1 t C e c) Show mathematically that the rate at which the energy is being removed from the capacitor is eual to the power instantaneously dissipated in the resistor, at any time t > We can take the derivative of U(t) with respect to time to get the rate of change of the stored energy Since we re interested in the rate at which energy is being removed from the capacitor, this uantity will be negative: du = d t dt dt C e = C = t e The power instantaneously dissipated in the resistor is I R, and we can also solve for this based on the known I(t) for a discharging capacitor: e t I ( t) = I e t = P R = I R = which is the exact same expression we had before t e t e R = t e,

5 Physical Sciences 3 Feb 7 1, 13 4 Pre-Lab uestion for Lab ( pts) Please solve this problem before coming to Lab, although you do not need to hand it in separately A schematic circuit diagram of a real voltage source (battery) is shown below The real voltage source can be modeled as an ideal voltage source along with an internal resistance r in series; the two cannot be separated (ie you can t poke around inside the dashed line with a voltmeter) However, you can calculate the internal resistance along with the EMF of the ideal voltage source by connecting an external resistor R, as shown a) You want to measure the voltage across R Draw a circuit diagram that includes the original circuit, and then add the voltmeter to the diagram in the appropriate place with the correct connections You can represent the voltmeter by a V with a circle around it The voltmeter needs to be wired in parallel with R in order to correctly measure the voltage across R (See figure to right) b) Determine an expression for the voltage across the external resistor R, in terms of E, R, and r We can use Kirchhoff s loop rule to write ΔV R in terms of E, R, and r only Note that we are not allowed to include I (the current) in our answer: The voltage across the external resistor is eual to ε IR Ir = I = ΔV R = IR = ε R + r ε R + r ( R 5