Download FREQUENTLY ASKED QUESTIONS Content

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?
. If power P is constant, you can find
Power is energy per time: P = dE
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
Rt
power over time: E = t0f 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.
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 28-4 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
charged-up 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.
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 newly-connected 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) = Q0 e−t/RC , where Q0 is the original stored charge. The
Q −t/RC
e
(negative sign just
current is the time derivative of that, I(t) = − RC
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 28-14?
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
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 RA and
RB in series with the battery, with a common current IAB through them. If
RC is connected by the switch, that is putting a new resistor RC in parallel
with RA and RB . This decreases the equivalent resistance of the circuit, so
there will be more total current pumped by the battery. But RA and RB
have the same potential difference across them as before (the battery’s ε).
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, Iε,
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, ε = Ir + IRAB , where RAB = RA + RB ,
the series resistance of RA and RB . This gives I = r+Rε AB , and a potential
difference across RA and RB of ∆VABbef = IRAB = r+Rε AB RAB . (BTW this
situation is called a “voltage divider”).
Now when RC is connected in parallel with RA and RB , we get equivalent
RAB
resistance RABC = RRCC+R
. We can then find the potential difference across
AB
this parallel network as above, from ∆VABC = ∆VC = ∆VAB = r+RεABC RABC .
Now because we know that RABC < RAB (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 ∆VABC < ∆VABbef (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 ∆VAB = r+RεABC RABC . (v) Total power delivered, Iε,
increases, because total I increases.
In question 28-39, why are Icap and Ibatt 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 left-hand branch with the battery is Ibatt . The loop
rule gives that ε − Ibatt R1 = 0. So Ibatt = ε/R1 : 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 right-hand branch with the capacitor. This has current
Icap , The right-hand loop is now exactly the same as a standard capacitor
discharge loop with resistance R2 and capacitance C, and its behavior is
identical to the one described in section 28.4 of your text: the current is
I = RQ2 C e−t/(R2 C) : R2 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 Ibatt +Icap , 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!