
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.


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.

 2 Ω is in series with
6 Ω. 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 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.

