Physics 2102 Jonathan Dowling. b. a. Physics 2102 Lecture 11. DC Circuits. Incandescent light bulbs. Which light bulb has a smaller resistance: a 60W, or a 100W one? Is the resistance of a light bulb different when it is on and off?
d=aEMF devices and single loop circuits
The battery operates as a “pump” that moves
positive charges from lower to higher electric potential. A battery is an example of an “electromotive force” (EMF) device.
These come in various kinds, and all transform one source of energy into electrical energy. A battery uses chemical energy, a generator mechanical energy, a solar cell energy from light, etc.
The difference in potential energy that the
device establishes is called the EMF
and denoted by E.
E = iR
Given the emf devices and resistors in a circuit, we want to calculate the circulating currents. Circuit solving consists in “taking a walk” along the wires. As one “walks” through the circuit (in any direction) one needs to follow two rules:
When walking through an EMF, add +Eif you flow with the current or -E
otherwise. How to remember: current “gains” potential in a battery.
When walking through a resistor,add -iR, if flowing with the current or +iR
otherwise. How to remember: resistors are passive, current flows “potential down”.
Walking clockwise from a: + E-iR=0.
Walking counter-clockwise from a: - E+iR=0.
If one connects resistors of lower and lower value of R to get higher and higher currents, eventually a real battery fails to establish the potential difference E, and settles for a lower value.
One can represent a “real EMF device” as an ideal one attached to a resistor,
called “internal resistance” of the EMF device:
E –i r -i R=0 i=E/(r+R)
The true EMF is a function of current: the more current we want, the smaller Etruewe get.
Etrue = E –i r
An electrical cable consists of 100 strands of fine wire, each having 2 W resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 5 A.
Assume now that the same 2 W strands in the cable are tied in series, one after the other, and the 100 times longer cable connected to the same 0.1 Volts potential difference as before.
Two resistors are “in series” if they are connected such that the same current flows in both.
The “equivalent resistance” is a single imaginary resistor that can replace the resistances in series.
“Walking the loop” results in :E –iR1-iR2-iR3=0 i=E/(R1+R2+R3)
In the circuit with the equivalent resistance, E –iReq=0 i=E/Req
Two resistors are “in parallel” if they are connected such that there is the same potential drop through both.
The “equivalent resistance” is a single imaginary resistor that can replace the resistances in parallel.
“Walking the loops” results in :E –i1R1=0, E –i2R2=0, E –i3R3=0
The total current delivered by the battery is i = i1+i2+i3 = E/R1+ E/R2+ E/R3.
In the circuit with the equivalent resistor,
Key formula: V=iR Q=CV
In series: same current same charge
In parallel: same voltage same voltage
Find the equivalent resistance between points
(a) F and H and
(b) F and G.
(Hint: For each pair of points, imagine that a battery is connected across the pair.)
If all capacitors have a capacitance of 6mF, and all batteries are ideal and have an emf of 10V, what is the charge on capacitor C?
If all resistors have a resistance of 4W, and all batteries are ideal and have an emf of 4V, what is the current through R?