
 Electric Circuits  Definition of terms
 Electric current I equals the rate of flow of current.
For constant current I = q/t. For a variable current
I = dq/dt.
 In Fig. 1 above, an electron moves in a conductor with the
equivalent of a constant drift velocity v_{d} in a
constant electric field E. In reality the electron experiences
a series of accelerations in the electric field after collisions
with nucleus and inner electrons of the atoms that make up
the material. Let n = the number of electrons/volume, e the
charge on the electron, A the crosssectional area of the
conductor, and L its length = v_{d}t, where
t is the time for the electron to move the length of the conductor.
The net charge in motion is then q = neAv_{d}t
and the current I = q/t = neAv_{d}. The current density
J = I/A = nev_{d}.
 Sample Problem in 108
Problem Set for Circuits: Problem 1.
 Resistances
 The resistance R of a conductor is directly proportional
to its length L, inversely proportional to its crosssectional
area A, and depends on the material's resistively ρ.
R = ρL/A.
 Ohm's Law. The potential difference V_{ab}
across a resistance R from a to b with a current I
from a to b is given by Ohm's Law, an Empirical law,
V_{ab} = IR.
 Combinations of Resistors
 Resistors in Series
 The current I is the same in each resistor
 The potential difference across all of the resistors
equals the sum of the individual potential differences:
V_{ab} = V_{ac} + V_{cd}
+ V_{db}
 R_{eq} = V_{ab}/I = V_{ac}/I
+ V_{cd}/I + V_{db}/I = R_{1 }+
R_{2 } + R_{3
} (Fig. 2a below)
 Resistors in Parallel
 The potential difference V_{ab } across
each of the resistors is the same.
 The total current I equals the sum of the currents
in each of the resistors:
 I = I_{1 }+ I_{2 } + I_{3
}
 I/V_{ab} = I_{1}/V_{ab }+
I_{2}/V_{ab } + I_{3}/V_{ab
}
 1/R_{eq }= 1/R_{1 } + 1/R_{2
} + 1/R_{3 } (Fig.
2b below)
 Sample Problems in 108
Problem Set for Circuits: Problems 2, 3,
6 and 7.
 Power
 Power = P = (U_{a } U_{b})/t _{ }=
q(V_{a } V_{b })/t = (q/t)(V_{a }
V_{b })_{ }= IV_{ab
} The above statement is always true.
 If the circuit element between a and b is an Ohmic
resistor for which
V_{ab }= IR, then P = I(V_{ab }) can
be written as P = I(IR) = I^{2}R or
P = (I)V_{ab } can be written as P = (V_{ab}/R)V_{ab}
= (V_{ab})^{2}/R.
 Sample Problem in 108
Problem Set for Circuits: Problem 4.
 Electromotive force ε
 The electromotive force ε =
energy/charge = U/q or
q ε = U and power
= P = U/t = (q/t) ε = I ε.
 For the circuit of Fig. 3 above, r is the internal resistance
of the battery. For this circuit:
Power in = Power out
Chemical power to electrical power = electrical power to
heat
I ε = I^{2}R
+ I^{2}r (Equation
1)
 From Eq. 1,
 I = ε/(R + r)
 ε  Ir
= IR = V_{ab
}
 The terminal voltage across a battery delivering
energy to the rest of the circuit, V_{ab}
= ε  Ir.
V_{ab} = ε + Ir
for a source of emf receiving
energy from the rest of the circuit.
 Sample Problems in 108
Problem Set for Circuits: Problems 5 and 8.
 Kirchhoff's Rules
 Junction rule. The sum of the currents entering any junction
equals the sum of the currents leaving that junction. This
is just a statement of conservation of charge.
 Loop rule. The sum of the potential differences across each
element around any closed circuit loop is zero. This is just
a statement of conservation of energy.
 If you go from the minus to the plus side of the battery,
chemical energy is being changed into electrical energy
and you are picking up energy, thus the sign is plus.
If you go from the plus side of the battery to the minus
sign, the sign is negative.
 If go around the loop in the same direction as the current, electrical
energy is being converted to heat and you are losing energy
so this IR is negative. If you go around the loop opposite
to the direction of the current, you are going from a
lower potential to a higher potential so this IR is positive.
 General comment. Once you choose directions of currents,
and going around the loop clockwise or counterclockwise,
stay with it. You may reverse the direction of going
around the loop for a second loop, but don't change the
directions of the currents.
 Sample problem: 108
Problem Set for Circuits: Problems 9 and 12.
 RCCircuits
 The potential difference across an Ohmic resistor = IR
and across a capacitor = q/C
 When the switch S is thrown up for the charging position
 Using the loop rule, we find
ε = IR
+ q/C.
Since I = dq/dt, we may write this as
ε = dq/dt
R + q/C (Equation
2)
Arithmetic gives us
(dq/dt)RC = (q  εC) (Equation
3)
Separating the variables:
dq/(q  εC)
= 1/RC dt (Equation
4)
Integrating both sides:
or
ln [(q εC)/ εC}
= t/RC
(Equation
5)
Taking the antilog of both sides of Eq. 5:
(q εC)/ εC
= e^{t/RC}
q  εC =  εC(e^{t/RC })
or
finally
q(t) = εC(1  e^{t/RC })
(Equation
6)
 The potential difference across the capacitor for charging
the capacitor V_{cb} sometimes called
V_{C } = q/C = ε(1
 e^{t/RC }).
(Equation
7)
 The current I = dq/dt = ε/R(
e^{t/RC} )
(Equation
8)
 The potential difference across the resistor for charging
the capacitor V_{ac} sometimes called
V_{R }= IR = ε(
e^{t/RC }) (Equation
9)
 For Figures 4 (above) and 5 (below), I have chosen
ε = 1.0V,
R = 10^{6} Ω
= 10^{6 }V/A, and C = 5 x 10^{6 }F
= 5 x 10^{6 }C/V. The time constant = RC
=10^{6 }V/A x 5 x 10^{6 }C/V = 5 C/A
= 5 C/(C/s) = 5 s.
For build up of charge, V_{C}(t) = ε(1
 e^{t/RC}). When t = RC = 5 s,
V_{C}(RC) = 1.0 V(1  e^{RC/RC} ) = 1.0
V( 1  e^{1}) = 1.0 V(0.632) = 0.632 V, as shown
below.
For build up of charge, V_{R}(t) = εe^{t/RC}.
When t = RC = 5 s,
V_{C}(RC) = 1.0 V( e^{RC/RC} ) = 1.0
V(e^{1}) = 1.0 V(0.368) = 0.368 V, as shown above.
 When the switch is thrown down, the battery is no longer
in the circuit and the capacitor discharges.
 V_{ab} = 0 = IR + q/C = dq/dt R + q/C (Equation
10)
Rearranging the equation: dq/q = dt/RC
The limits are now from maximum charge Q = εC
to q when t goes from 0 to t. The capacitor is discharging
and I is in the opposite direction.
ln q/ εC =  t/RC
or
q(t) = Qe^{t/RC} = εCe^{t/RC
} (Equation
11)
 I(t) = dq/dt = ( εC/RC)e^{t/RC}
=  ( ε/R)e^{t/RC
}
 V_{cb } = V_{C } = q/C = εe^{t/RC}
and
V_{ac} = IR =  εe^{t/RC
}
 Referring to Fig. 5b below,
For decay of charge, V_{C}(t) = ε
e^{t/RC}. When t = RC = 5 s,
V_{C}(RC) = 1.0 V(e^{RC/RC} ) = 1.0 V(
e^{1}) = 1.0 V(0.368) = 0.368 V.
For decay of charge, V_{R}(t) =  εe^{t/RC}.
When t = RC = 5 s,
V_{R}(RC) = 1.0 V(e^{RC/RC}) = 1.0
V(e^{1}) = 1.0 V(0.368) = 0.368 V.
 Practice Problems in 108
Problem Set for Circuits: Problems 10, 11,
and 13.

