
1.

 The average current density is
J = I/A = 20A/ π(1.63 x 10^{3} m)^{2}
= 2.4 x 10^{6} A/m^{2}.
 v_{d} = J/en = 2.4 x 10^{6} A/m^{2}/(1.6
x 10^{19}C)(8.47 x 10^{28 }m^{3})
=
1.8 x 10^{4 }m/s.
 E = ρJ = (1.56 x 10^{8} Vm/A)(2.4 x 10^{6}
A/m^{2}) = 0.037 V/m.
 R = ρL/A or
L = RA/ρ = 0.012 Ω (1.2 x 10^{6 }m^{2})/1.56
x 10^{8 }Ω  m = 0.92 m.


2.

If the current I_{1 }= 3.0 A in the 4.0 Ω
resistor,
V_{cb } = I_{1 }(4.0 Ω) = 3.0 A(4.0 Ω)
= 12 V.
I_{2 }= V_{cb}/6.0 Ω_{ }= 12
V/6.0 Ω = 2.0 A.
I = I_{1 }+ I_{2 }= 3.0 A + 2.0 A = 5.0 A.
V_{ac} = IR_{ac} = (5.0 A)(5.0 Ω) = 25
V.
V_{ab} = V_{ac} + V_{cb} = 25 V +
12 V = 37 V.


3.

 R_{c’e’} = (2 + 6) Ω= 8 Ω.
 3 Ω is in series with
the 21 Ω. R_{c”e”}
= (3 + 21) Ω = 24 Ω.
 R_{c’e’} is in parallel with R_{c”e”}.
1/R_{ce} = (1/8 + 1/24)Ω^{1}
= (3/24 + 1/24) Ω^{1}.
R_{ce} = 6 Ω.
 R_{ab} = R_{ac} + R_{ce} + R_{eb}
= (2 + 6 + 4) Ω = 12
Ω.
 I = V_{ab}/R_{ab} = 24 V/12 Ω
= 2 A.
 V_{ce} = IR_{ce} = 2A(6 Ω)
= 12 V = V_{c’e’ }= V_{c”e”}.
 I_{1 }= V_{c’e’}/R_{c’e’}
= 12 V/ 8 Ω = 3/2 A.
 I_{2} = V_{c”e”}/R_{c”e”
}= 12 V/24 Ω =
1/2 A. Note that I_{1} + I_{2}
= I.


4.

For the resistors in series, the current in each is the same
and the power
P = I^{2}R. Since R_{1} > R_{2},
P_{1 }> P_{2}.
When the resistors are in parallel, the current is not the same
in each, but the potential difference across each is the same.
P_{1} = (I_{1}’) V_{ab}
= (V_{ab}/R_{1})(V_{ab}) = V_{ab}^{2}/R_{1}.
P_{2} = (I_{2}’) V_{ab} = (V_{ab}/R_{2})(V_{ab})
= V_{ab}^{2}/R_{2}.
Since R_{1} > R_{2}, now P_{2}
> P_{1}.


5.

 The total resistance of the circuit is (9.5 + 20 + 0.5)Ω
= 30 Ω.
The current I = 30 V/30 Ω=
1 A.
 Power dissipated in external resistances =
I^{2}(9.5 + 20)Ω
= 1 A^{2 }(29.5 V/A) = 29.5 W.
Power dissipated in the internal resistance =
I^{2}(0.5 Ω)
= (1 A^{2}) (0.5 V/A) = 0.5 W.
Total power dissipated in resistances =
(29.5 + 0.5) W = 30 W.
 Power supplied when chemical energy is changed into electrical
energy = (Emf)(I) = 30 V (1 A) = 30 W.


6.

 When both switches are closed, the resistance of the upper
branch of the parallel circuit is (2 + 1) Ω =
3Ω and that of
the lower branch is
(2 + 4) Ω = 6Ω.
For the parallel group, 1/R_{eq} = 1/3 Ω +
1/6 Ω and
R_{eq} = 2Ω.
The total resistance of the circuit is (2 + 4) Ω
= 6 Ω and
I = 12 V/6 Ω = 2 A.
 If only switch A is closed, the bottom part of the parallel
grouping is not in the circuit. The total resistance of
the circuit is (2 + 1 + 4) Ω
and
I = 12/7 A.
 If only switch B is thrown, the top part of the parallel
grouping is not in the circuit. The total resistance of
the circuit is (2 + 4 + 4) Ω and
I = 12V/10 Ω = 1.2 A.


7.

In the Fig. for #7, the 10 Ω
resistance is wired in parallel with the 15 Ω
resistance from c to b’. This parallel grouping is then
wired in series with 4.0 Ω
resistor between a’ and b’. Finally this grouping
is wired in parallel with another 10 Ω
resistor.
1/R_{cb’} = 1/10 Ω
+ 1/15 Ω = (3 +2)/30 Ω.
R_{cb’ }= 6 Ω.
R_{ab’ }= (6 + 4) Ω
= 10 Ω.
1/R_{ab} = 1/10 Ω
+ 1/10 Ω = 2/10 Ω.
R_{ab} = 5.0 Ω


8.

For the circuit in Fig. for #8 above, I= ε/(r
+ R).
For I = 6.0 A and R = 2.0 Ω,
6.0 A = ε/(r
+ 2.0 Ω)
(Equation
1)
For I = 3.0 A and R = 7.0 Ω,
3.0 A = ε/(r
+ 7.0 Ω)
(Equation
2)
From Eq. 1: ε = 6.0A
r + 12 V. From Eq. 2: ε
= 3.0A r + 21V.
Thus, 6.0A r + 12 V = 3.0A r + 21 V. r = 9.0V/3.0
A = 3.0 Ω.
Substituting r back into Eq. 1, 6.0 A = ε/(3.0
+ 2.0) Ω. ε
= 30 V.


9.

From conservation of charge,
I_{1 }= I_{2} + I_{3} or
I_{3 }= I_{1 } I_{2}
(Equation
1)
For loop I,
36 V  I_{1}(0.5 + 2 + 1.5) Ω 6 V  I_{2}(3 Ω)
= 0 or
30 A = 4I_{1} + 3I_{2}
(Equation
2)
For loop II,
6 V + (3 Ω)I_{2}
 4 V  (2 Ω)I_{3 }=
0 or
2A = 2I_{3}  3I_{2 }(Equation
3)
Substitute Eq. (1) into Eq. (3):
2A = 2(I_{1}  I_{2}) 3I_{2}
or
2A = 2I_{1}  5I_{2}
(Equation
4)
2 x Eq. 4 equals: 4A
= 4I_{1}  10I_{2}
(Equation
5)
Eq. 2  Eq. 5 equals: 26A
= 13I_{2} or
I_{2 }= 2A.
Then from Eq. 2: 30A
= 4I_{1} + 3(2A) and
I_{1} = 6 A.
From Eq. 1, I_{3} = 6A  2A = 4A.


10.

Given q(t) = Q_{i} e^{t/RC}. When t =
RC, q (RC) = Q_{i} e ^{RC/RC} = Q_{i
}e^{1 }= Q_{i}/e.
The time constant for the circuit is RC.
 In Fig. 6 above, the two capacitors are in parallel. For
capacitors in parallel, the equivalent capacitance is the
sum of the capacitances.
C_{ab }= (10 + 10)µF = 20µF
= 20 x 10^{6 }C/V.
The two resistances are in series. The equivalent resistance
of resistances in series is the sum of the resistances.
R_{bd} = (5.0 + 2.0)MΩ=
7.0 MΩ = 7.0 x 10^{6
}V/A.
The time constant =
RC = 7.0 x 10^{6 }V/A x 20 x 10^{6
}C/V = 140 C/A = 140 s.
 q(t) = Q_{i}e^{t/RC}. For Q_{i
}= 100 µC and q = 50µC, q/Q_{i }
= 1/2 = e^{t/RC
} or
e^{t/RC} = 2. Taking log of both sides of
this equation, t/RC = ln 2.
t = RC ln 2 = 140 s (0.693) = 97 s.


11.

 V_{ab }= V_{ac }+ V_{cb
} ε = q/C + RI =
q/C + Rdq/dt.
q  εC =  RC dq/dt,
or
separating variables
dq/(q  εC) =  dt/RC.
Integrating,
_{Qo }∫^{q }dq/(q  εC)
= 1/RC _{o}∫^{t} dt
ln (q  εC)/(Q_{o}
 εC) =  t/RC.
Taking the antilog of both sides of the equation and rearranging,
q(t) = Q_{o} e^{t/RC} + εC(1
 e^{t/RC }).
I(t) = dq/dt =  Q_{o}/RC e^{t/RC} +
ε/R e^{t/RC}.
q(0) = Q_{o }= C(V_{ac})_{o} =
4.0 x 10^{6 }C/V(100 V) = 4.0 x 10^{4}
C.
I(0) =  Q_{o}/RC + ε/R
= (100 + 40)V/500 Ω
= 0.12 A.
 q(∞) = εC =
40 V(4.0 x 10^{6} C/V) = 1.6 x 10^{4}C.
I(∞) = 0.


12.

The current in the circuit of Fig. for #12a above, I =
ε/(r + R), and the power
delivered to the load resistor R, is P = I^{2}R = ε^{2}R/(r
+ R)^{2}.
For maximum power,
dP/dR = ε^{2}[(r
+ R)^{2} 2R(r + R)]/(r + R)^{4} = 0,
so
[(r + R)^{2} 2R(r + R)] = 0 or
r + R = 2R and
R = r.
Again P = ε^{2}R/(r
+ R)^{2}. For R = 0, P = 0. As R approaches
a very large number, P approaches ε/R
which goes to zero. In the Fig. for #12 b above, I have
taken ε = 2.0 V and
r = 0.5 Ω. Notice
that P is a maximum at R = r = 0.5 Ω.


13.

 Just after the switch is thrown, the capacitor has no
charge. It is just as though there were a short in the circuit
or replacing the capacitor by a wire. Then the two resistance
between points c and b are in parallel.
(See Fig. 8b below).
1/R_{cb} = 1/R + 1/R = 2/R and R_{cb} =
R/2. This equivalent resistance R_{cb} is in series
with the resistance R between a and c. (Fig. 8c below).
The total resistance of the circuit =
R_{AB} = R + R/2 = 3R/2 = 3(2000 Ω/2)
= 3000 Ω (Fig.
8d below).
I_{1 }= ε/R_{ab
} = 6 V/3000 Ω =
2 x 10^{3 }A.
V_{ac} = I_{1}(R/2) = 2 x 10^{3}
A(1000 Ω) = 2 V.
I_{2} = I_{3 }= V_{ac}/R = 2 V/2000
Ω = 1 x 10^{3
}A.
Notice that I_{1} = I_{2} + I_{3}.
 After the switch has been thrown for a very long time,
the capacitor is fully charged and acts like an open circuit.
The circuit reduces to that shown in Fig. 8e below.
I_{3} = 0 and
I_{1} = I_{2} = I = ε/2R
= 6 V/2000 Ω = 3 x 10^{3}A.
From conservation of charge,
I_{1} = I_{2} + I_{3} (Equation
1)
From conservation of energy,
Loop I: ε
 I_{1}R – I_{2}R = 0
(Equation
2)
Loop II: I_{2}R –
I_{3}R  q/C = 0 (Equation
3)
Substituting Eq. 1 into Eq. 2:
ε  (I_{2
}+ I_{3})R – I_{2}R = 0
(Equation
4)
Solving for I_{2} in Eq. 4:
I_{2} = ( ε
 I_{3}R)/2R
(Equation
5)
Substituting Eq. 5 into Eq. 3:
( ε  I_{3}R)/2
– I_{3}R + q/C = 0
(Equation
6)
Rearranging Eq. 6:
ε = 3I_{3}R
+ 2q/C (Equation
7)
Recognizing that I_{3} = dq/dt, Eq. 7 becomes
:
ε = 3(dq/dt)R
+ 2q/C (Equation8)
The equation we solved for in the Outline for Circuits was
ε = (dq/dt)R
+ q/C (Equation
9)
and had solution q(t) = C ε(1
– e^{t/RC}) and time constant τ
= RC.
Comparing Eq. 8 and Eq. 9, we see for Eq. 8 that ε
becomes ε/2
and the time constant τ
= 3RC/2.
For this case,
q(t) = (C ε/2)(1
– e^{2t/3RC}) and
I_{3}(t) = dq/dt = ( ε/3R)
e^{2t/3RC} (Equation
10)
I_{3}(0) = ( ε/3R)
e^{0} = ( ε/3R)
and
I_{3}(∞) = 0, as in Part (b) above.
From Eq. 5,
I_{2} = ( ε
 I_{3}R)/2R = ε/2R
 I_{3}/2 = ε/2R
 ( ε/6R) e^{2t/3RC
}I_{2}(t) = ε/6R(3
 e^{2t/3RC}) (Equation
11)
I_{2}(0) = ε/3R
and
I_{2}(∞) = ε/2R,
as in Part (b) above.
From Eq. (1),
I_{1} = I_{2} + I_{3 }=
( ε/6R)(3  e^{2t/3RC})
+ ( ε/3R) e^{2t/3RC}
I_{1}(t) = ( ε/6R)
(3 + e^{2t/3RC})
(Equation
12)
I_{1}(0) = (2 ε/3R)
and
I_{1}(∞) = ( ε/2R),
as in Part (b) above.
A plot of I_{1}, I_{2}, and I_{3
}as a function of time is shown in Fig. 8g below.

