Lecture 12 - Circuits
Chapter 31 - Tuesday February 20th
•DC circuits and electromotive force
•Kirchoff’s 1st and 2nd laws (all you need to know!)
•Analysis of circuits
•Internal resistance and emf
•Energy transfer in electric circuits
Reading: pages 701 thru 716 (chapter 31) in HRK
Read and understand the sample problems
WebAssign deadline will be Friday 23rd at 11:59pm
Graded problems (Ch.31) –E. 13, 25, 44, 49; P. 4, 8
Practice problems (Ch. 31): Ex. 25, 39, 47, 49
DC Circuits
emf
DC Circuits
Kirchoff’s first law:
At any junction in an electrical circuit, the total current entering
the junction must equal the total current leaving the junction.
Electromotive force (emf)
•Source of electrical energy in a circuit.
E = dW/dq
SI unit: joule/Coulomb
•Represents the potential energy provided to each coulomb
of charge that passes through the device.
•IT IS NOT A FORCE!!!
•Most often, emf is provided by a battery (a chemical cell).
•The emf is the same as the potential difference between
the negative and positive terminals of a battery WHEN NO
CURRENT FLOWS.
•In general, when a current flows, the potential difference
at the terminals of a battery is lower than the emf.
•An emf can also store energy.
Circuit analysis
Kirchoff’s second law:
The algebraic sum of all differences in potential
around a complete circuit loop must be zero.
Internal resistance
i=
ε
R +r
ΔVab =
ε
R
R +r
Electric fields
G G
W = ∫ F ⋅ ds
G
W
F G
E = = ∫ ⋅ ds
q
q
V
G
G G
F G
v∫ q ⋅ d s = E + v∫ E ⋅ d s = E
Electric fields
j = σE
Energy transfer in electric circuits
•A 1V battery does work by providing each coulomb of
charge that leaves its positive terminal 1 joule of energy.
•If charge flows at a rate of 1 coulomb per second, then
the battery does work at a rate of 1 joule per second,
i.e.
joule
coulomb
joule
Power=
×
=
=watt
coulomb second
second
P = Ei = dW/dt
•In a resistor, energy is lost in an amount iR per coulomb.
⇒ Pcharge = iΔV = i(−iR) = −i2R
Pheat = i2R