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Phyllis Fleming Physics
Physics 108
Answers - Circuits
1.
a.
The average current density is
b.
vd = J/en = 2.4 x 106 A/m2/(1.6 x 10-19C)(8.47 x 1028 m-3) =
1.8 x 10-4 m/s.
c.
E=
d.
R=
J = I/A = 20A/(1.63 x 10-3 m)2 = 2.4 x 106 A/m2.
J = (1.56 x 10-8 V-m/A)(2.4 x 106 A/m2) = 0.037 V/m.
L/A or
L = RA/ = 0.012  (1.2 x 10-6 m2)/1.56 x 10-8  - m = 0.92 m.
2.
If the current I1 = 3.0 A in the 4.0
 resistor,
) = 3.0 A(4.0 ) = 12 V.
I2 = Vcb/6.0  = 12 V/6.0  = 2.0 A.
Vcb = I1 (4.0
I = I1 + I2 = 3.0 A + 2.0 A = 5.0 A.
Vac = IRac = (5.0 A)(5.0 ) = 25 V.
Vab = Vac + Vcb = 25 V + 12 V = 37 V.
3.
= 8 .
a.
Rc’e’ = (2 + 6)
b.
3
c.
Rc’e’ is in parallel with Rc”e”.
 is in series with the 21 .
Rc”e” = (3 + 21)
1/Rce = (1/8 + 1/24)-1 = (3/24 + 1/24)
d.
Rab = Rac + Rce + Reb = (2 + 6 + 4)
e.
I = Vab/Rab = 24 V/12
f.
Vce = IRce = 2A(6
g.
I1 = Vc’e’/Rc’e’ = 12 V/ 8
h.
I2 = Vc”e”/Rc”e” = 12 V/24
-1.
 = 24 .
Rce = 6
.
 = 12 .
 = 2 A.
) = 12 V = Vc’e’ = Vc”e”.
 = 3/2 A.
 = 1/2 A.
Note that I1 + I2 = I.
4.
For the resistors in series, the current in each is the same and the power
P = I2R. Since R1 > R2, P1 > P2.
When the resistors are in parallel, the current is not the same in each, but the
potential difference across each is the same.
P1 = (I1’) Vab = (Vab/R1)(Vab) = Vab2/R1.
P2 = (I2’) Vab = (Vab/R2)(Vab) = Vab2/R2.
Since R1 > R2, now P2 > P1.
5.
a.
The total resistance of the circuit is (9.5 + 20 + 0.5)  = 30
The current I = 30 V/30
b.
= 1 A.
Power dissipated in external resistances =
.
I2(9.5 + 20) = 1 A2 (29.5 V/A) = 29.5 W.
Power dissipated in the internal resistance =
I2(0.5
) = (1 A2) (0.5 V/A) = 0.5 W.
Total power dissipated in resistances =
(29.5 + 0.5) W = 30 W.
c.
Power supplied when chemical energy is changed into electrical energy =
(Emf)(I) = 30 V (1 A) = 30 W.
a.
When both switches are closed, the resistance of the upper branch of the
6.
parallel circuit is (2 + 1)
 = 3
and that of the lower branch is
(2 + 4) = 6.
For the parallel group, 1/Req = 1/3
 + 1/6 
The total resistance of the circuit is (2 + 4)
I = 12 V/6 = 2 A.
b.
and Req = 2.
=6
and
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)
I = 12/7 A.
c.
and
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)
I = 12V/10
7.

 = 1.2 A.
and
 resistance is wired in parallel with the 15  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 
In the Fig. for #7, the 10
resistor.
1/Rcb’ = 1/10
Rcb’ = 6
.
 + 1/15  = (3 +2)/30 .
 = 10 .
1/Rab = 1/10 + 1/10  = 2/10.
Rab = 5.0 
Rab’ = (6 + 4)
8.
For the circuit in Fig. for #8 above,
For I = 6.0 A and R = 2.0
6.0 A = /(r + 2.0
)
,
(Equation 1)
For I = 3.0 A and R = 7.0,
3.0 A = /(r + 7.0
I= /(r + R).
) (Equation 2)
 = 6.0A r + 12 V. From Eq. 2:  = 3.0A r + 21V.
Thus, 6.0A r + 12 V = 3.0A r + 21 V. r = 9.0V/3.0 A = 3.0.
Substituting r back into Eq. 1, 6.0 A = /(3.0 + 2.0).  = 30 V.
From Eq. 1:
9.
From conservation of charge,
I1 = I2 + I3 or I3 = I1 - I2 (Equation 1)
For loop I,
36 V - I1(0.5 + 2 + 1.5)- 6 V - I2(3
30 A = 4I1 + 3I2
(Equation 2)
For loop II,
) = 0 or
6 V + (3)I2 - 4 V - (2 )I3 = 0 or
2A = 2I3 - 3I2 (Equation 3)
Substitute Eq. (1) into Eq. (3):
2A = 2(I1 - I2) -3I2 or 2A = 2I1 - 5I2
(Equation 4)
2 x Eq. 4 equals:
4A = 4I1 - 10I2
(Equation 5)
Eq. 2 - Eq. 5 equals:
26A = 13I2
Then from Eq. 2:
30A = 4I1 + 3(2A)
From Eq. 1,
or
I2 = 2A.
and I1 = 6 A.
I3 = 6A - 2A = 4A.
10.
Given q(t) = Qi e-t/RC. When t = RC, q (RC) = Qi e
The time constant for the circuit is RC.
a.
-RC/RC
= Qi e-1 = Qi/e.
In Fig. 6 above, the two capacitors are in parallel. For capacitors in parallel,
the equivalent capacitance is the sum of the capacitances.
Cab = (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.
Rbd = (5.0 + 2.0)M= 7.0 M = 7.0 x 106 V/A.
The time constant =
RC = 7.0 x 106 V/A x 20 x 10-6 C/V = 140 C/A = 140 s.
b.
q(t) = Qie-t/RC. For Qi = 100 µC and q = 50µC, q/Qi = 1/2 = e-t/RC
or et/RC = 2. Taking log of both sides of this equation, t/RC = ln 2.
t = RC ln 2 = 140 s (0.693) = 97 s.
a.
Vab = Vac + Vcb
11.
 = q/C + RI = q/C + Rdq/dt.
q - C = - RC dq/dt, or separating variables
dq/(q - C) = - dt/RC.
Integrating,
Qo
∫q dq/(q - C) = -1/RC o∫t dt
ln (q - C)/(Qo - C) = - t/RC.
Taking the antilog of both sides of the equation and rearranging,
q(t) = Qo e-t/RC + C(1 - e-t/RC ).
I(t) = dq/dt = - Qo/RC e-t/RC + /R e-t/RC.
q(0) = Qo = C(Vac)o = 4.0 x 10-6 C/V(100 V) = 4.0 x 10-4 C.
I(0) = - Qo/RC + /R = (-100 + 40)V/500
b.
 = 0.12 A.
q(∞) = C = 40 V(4.0 x 10-6 C/V) = 1.6 x 10-4C.
I(∞) = 0.
12.
The current in the circuit of Fig. for #12a above, I = /(r + R), and the power
delivered to the load resistor R, is P = I2R =
For maximum power,
dP/dR =
2R/(r + R)2.
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.
2R/(r + R)2. For R = 0, P = 0. As R approaches a very large number, P
approaches /R which goes to zero. In the Fig. for #12 b above, I have taken  =
2.0 V and r = 0.5 . Notice that P is a maximum at R = r = 0.5 .
Again P =
13.
a.
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/Rcb = 1/R + 1/R = 2/R and Rcb = R/2. This equivalent resistance Rcb is in
series with the resistance R between a and c. (Fig. 8c below).
The total resistance of the circuit =
RAB = R + R/2 = 3R/2 = 3(2000
/2) = 3000 
(Fig. 8d below).
I1 = /Rab = 6 V/3000
 = 2 x 10-3 A.
Vac = I1(R/2) = 2 x 10-3 A(1000 ) = 2 V.
I2 = I3 = Vac/R = 2 V/2000  = 1 x 10-3 A.
Notice that I1 = I2 + I3.
b.
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.
I3 = 0 and I1 = I2 = I = /2R = 6 V/2000
 = 3 x 10-3A.
c.
From conservation of charge,
I1 = I2 + I3
(Equation 1)
From conservation of energy,
Loop I:
Loop II:
 - I1R – I2R = 0
I2R – I3R - q/C = 0
(Equation 2)
(Equation 3)
Substituting Eq. 1 into Eq. 2:
 - (I2 + I3)R – I2R = 0
(Equation 4)
Solving for I2 in Eq. 4:
I2 = ( - I3R)/2R
(Equation 5)
Substituting Eq. 5 into Eq. 3:
( - I3R)/2 – I3R + q/C = 0
(Equation 6)
Rearranging Eq. 6:
 = 3I3R + 2q/C
(Equation 7)
Recognizing that I3 = dq/dt, Eq. 7 becomes :
 = 3(dq/dt)R + 2q/C
(Equation8)
The equation we solved for in the Outline for Circuits was
 = (dq/dt)R + q/C
(Equation 9)
and had solution q(t) = C(1 – e-t/RC) and time constant
Comparing Eq. 8 and Eq. 9, we see for Eq. 8 that
and the time constant
 = 3RC/2.
 = RC.
 becomes /2
For this case,
q(t) = (C/2)(1 – e-2t/3RC) and
I3(t) = dq/dt = (/3R) e-2t/3RC
(Equation 10)
I3(0) = (/3R)
= (/3R) and
I3(∞) = 0, as in Part (b) above.
e0
From Eq. 5,
I2 = ( - I3R)/2R = /2R - I3/2 = /2R - (/6R) e-2t/3RC
I2(t) = /6R(3 - e-2t/3RC)
(Equation 11)
I2(0) = /3R and
I2(∞) = /2R, as in Part (b) above.
From Eq. (1),
I1 = I2 + I3 = (/6R)(3 - e-2t/3RC) + (/3R) e-2t/3RC
I1(t) = (/6R) (3 + e-2t/3RC)
(Equation 12)
I1(0) = (2/3R) and
I1(∞) = (/2R), as in Part (b) above.
A plot of I1, I2, and I3 as a function of time is shown in Fig. 8g below.
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Phyllis J. Fleming
August 8, 2002
April 30, 2003