FREQUENTLY ASKED QUESTIONS October 2, 2012


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1 FREQUENTLY ASKED QUESTIONS October 2, 2012 Content Questions Why do batteries require resistors? Well, I don t know if they require resistors, but if you connect a battery to any circuit, there will always be at least some resistance in the circuit, whether you want there to be or not. Even a wire has a resistance, albeit a small one. Batteries also have internal resistance the internal resistance of a battery may be small enough to neglect, but it s always there. How do you know when charge is conserved or when voltage is conserved? When a circuit is isolated from the surroundings, and disconnected from a battery, charge is conserved there is no way for the charge to escape. Potential difference is not necessarily constant in this situation, however. If a circuit is connected to a battery, the potential difference across the battery s terminals is kept constant (that s the battery s job). Charge on a capacitor is not necessarily conserved though. What s the connection between energy and power? Power is energy per time: P = de. If power P is constant, you can find dt energy for a fixed time interval t from P t. However if P is not constant (say, for a case when current is changing due to a charging or a discharging capacitor), then to find energy supplied or dissipated, you need to integrate power over time: E = t f t 0 P (t)dt. How do you know if a junction exists or just a continuation of wire? A junction is a connection point in a circuit, where wires touch and current can divide. However the circuit must be a complete loop for the connection point to be a junction (hence the term circuit ). If you have a connection wire just going to a stub (an open circuit), then there is no current in the stub and you don t count the connection point as a junction.
2 What is a branch? A branch is a segment of a circuit between two junctions. There is a single current going through a branch. How do you know when current travels through a switch? A switch connects its terminals with a conductor when the switch is closed. So when a switch is closed (assuming it s part of an actual circuit), current can flow. When the switch is open, that s an open circuit and current can t flow because there is no conductor present. How do you derive the charging and discharging equations? These are derived using Kirchoff s loop rule: section 284 does the derivation. Basically, you set up an RC circuit, either with a battery for a charging case, or with a conducting path between capacitor plates for a discharging case. Then you go around the loop setting the sum of all the potential differences to zero. Then using that I = dq/dt, you get a differential equation that you can solve for Q(t) and I(t). I don t think you ll be asked to do this derivation on an exam, but you need to know the principles it s based on. You should go through the derivation and understand where the equations come from. How do you know when to use the charging or discharging equations? If a battery (which enforces a potential difference) is newly connected to an RC circuit, then the capacitor will charge up. In this case the appropriate equations are the charging ones. If something happens to allow charge to flow back from one plate of a chargedup capacitor to the other, e.g. a switch is thrown to short out the battery, so that the battery no longer enforces a potential difference across the capacitor, then it s a discharging situation and the discharging equations are the ones that describe it.
3 Does the direction of the current change when the capacitor goes from charging to discharging? Yes. When a capacitor is charging, current flows towards the positive plate (as positive charge is added to that plate) and away from the negative plate. When the capacitor is discharging, current flows away from the positive and towards the negative plate, in the opposite direction. Why does current go to zero as t in a discharging circuit? Thinking physically: as the capacitor discharges, there s at first a big potential difference between the newlyconnected plates and there will be lot of current driven by that difference. But as time goes on, more and more positive charge gets reunited with negative charge, the potential difference between the plates decreases, and current decreases. At very long times, essentially all of the charge separation is gone, and the plates become neutral. There will no longer be any charge flow, so the current dies away to zero. Thinking in terms of equations: the equation describing charge as a function of time is Q(t) = Q 0 e t/rc, where Q 0 is the original stored charge. The current is the time derivative of that, I(t) = Q RC e t/rc (negative sign just meaning direction of current opposite to the charging direction). Mathematically, the current goes to zero at infinite time. What would happen if there were no resistor and you closed a switch and discharged a capacitor? Where would the energy go? Well, there s never actually zero resistance in any circuit, although it may be very small. Even highly conductive wires always have some resistance. If you have only a wire attached to a capacitor and close a switch, a large current will flow in the wire, and the energy will be dissipated according to I 2 R. In practice, for a large capacitance and/or small resistance, there could be enough heat to melt the wire, or even cause sparks or explosions (don t try it at home!) The energy has to go somewhere it could go into heat, light, or mechanical energy. Can you explain objective question 2814? Treat the light bulbs as resistors. They burn more brightly if current is larger: energy dissipated according to I 2 R will be converted (partly) to light, so the
4 bigger I for a fixed R, the brighter the bulb. In this question, the presence of the small internal resistance of the battery complicates the situation a bit. Let s start by assuming no internal resistance of the battery, which gives a slightly simpler situation. Before the switch is closed, you have R A and R B in series with the battery, with a common current I AB through them. If R C is connected by the switch, that is putting a new resistor R C in parallel with R A and R B. This decreases the equivalent resistance of the circuit, so there will be more total current pumped by the battery. But R A and R B have the same potential difference across them as before (the battery s ). So the current through them will be the same as before. So for no internal resistance of the batter the answers are: (i) Lamp B s brightness does not change because the current through it is the same. (ii) Lamp C turns on. (iii) Lamp A s potential difference does not change (same current). (iv) Lamp C gets the potential difference of the battery. (v) Total power delivered, I, increases, because total I increases. But now assume (as specified in the problem) that there is some internal resistance of the battery r. This can be treated as a resistor in series with the battery. Before the switch is closed, = Ir + IR AB, where R AB = R A + R B, the series resistance of R A and R B. This gives I = r+r AB, and a potential difference across R A and R B of V ABbef = IR AB = situation is called a voltage divider ). Now when R C is connected in parallel with R A and R B, we get equivalent resistance R ABC = R CR AB R C +R AB. We can then find the potential difference across r+r AB R AB. (BTW this this parallel network as above, from V ABC = V C = V AB = r+r ABC R ABC. Now because we know that R ABC < R AB (you get less overall resistance when adding more resistance in parallel you re adding more ways for the charge to get through), we get that V ABC < V ABbef (in a voltage divider, the smaller the resistance, the smaller the fractional voltage drop across it). This leads to answers: (i) Lamp B s brightness goes down, because the voltage across it goes down and hence the current through it goes down. (ii) Lamp C turns on. (iii) Lamp A s potential difference goes down. (iv) Lamp C gets the potential difference V AB = r+r ABC R ABC. (v) Total power delivered, I, increases, because total I increases.
5 In question 2839, why are I cap and I batt independent when the switch is closed? Kirchoff s rules apply here. When the switch is closed, each branch has its own current, and each loop in the circuit can be treated with Kirchoff s loop rule. The current in the lefthand branch with the battery is I batt. The loop rule gives that I batt R 1 = 0. So I batt = /R 1 : so current acts as if it s just a battery plus resistor loop, regardless of whether or not there s extra current in the switch branch. The direction of the current is clockwise. Now consider the righthand branch with the capacitor. This has current I cap, The righthand loop is now exactly the same as a standard capacitor discharge loop with resistance R 2 and capacitance C, and its behavior is identical to the one described in section 28.4 of your text: the current is I = Q R 2 C e t/(r 2C) : R 2 is the relevant resistor. The direction of the current is counterclockwise. Now, Kirchoff s junction rule applies at the junctions on each side of the switch branch: total current in at the top junction is I batt +I cap, which equals total current in the switch. Jokes of the Week Q: What is the name of the first electricity detective? A. Sherlock Ω. Q: What s the difference between a battery and (insert whatever here)? A. A battery has a positive side!
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