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Physics 108

Answers - Circuits

1.
  1. The average current density is
    J = I/A = 20A/p(1.63 x 10-3 m)2 = 2.4 x 106 A/m2.

  2. 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.

  3. E = rJ = (1.56 x 10-8 V-m/A)(2.4 x 106 A/m2) = 0.037 V/m.

  4. R = rL/A   or
    L = RA/r = 0.012 W (1.2 x 10-6 m2)/1.56 x 10-8 W - m = 0.92 m.


2.



If the current  I1 = 3.0 A in the 4.0 W resistor,

Vcb = I1 (4.0 W) = 3.0 A(4.0 W) = 12 V.
I2 = Vcb/6.0 W = 12 V/6.0 W = 2.0 A.
I = I1 + I2 = 3.0 A + 2.0 A = 5.0 A.
Vac = IRac = (5.0 A)(5.0 W) = 25 V.
Vab = Vac + Vcb = 25 V + 12 V = 37 V.

3.
  1. Rc’e’ = (2 + 6) W = 8 W.

  2. 3 W is in series with the 21 W.  Rc”e” = (3 + 21) W = 24 W.

  3. Rc’e’ is in parallel with Rc”e”.
    1/Rce = (1/8 + 1/24)W-1 = (3/24 + 1/24) W-1.   Rce = 6 W.

  4. Rab = Rac + Rce + Reb = (2 + 6 + 4) W = 12 W.

  5. I = Vab/Rab = 24 V/12 W = 2 A.

  6. Vce = IRce = 2A(6 W) = 12 V = Vc’e’ = Vc”e”.

  7. I1 = Vc’e’/Rc’e’ = 12 V/ 8 W = 3/2 A.

  8. I2 = Vc”e”/Rc”e” = 12 V/24 W = 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.


  1. The total resistance of the circuit is (9.5 + 20 + 0.5)W = 30 W.
    The current  I = 30 V/30 W= 1 A.

  2. Power dissipated in external resistances =
    I2(9.5 + 20)W = 1 A2 (29.5 V/A) = 29.5 W.
    Power dissipated in the internal resistance =
    I2(0.5 W) = (1 A2) (0.5 V/A) = 0.5 W.
    Total power dissipated in resistances =
    (29.5 + 0.5) W = 30 W.

  3. Power supplied when chemical energy is changed into electrical energy = (Emf)(I) = 30 V (1 A) = 30 W.

6.

  1. When both switches are closed, the resistance of the upper branch of the parallel circuit is (2 + 1) W = 3W  and that of the lower branch is
    (2 + 4) W = 6W.

    For the parallel group,  1/Req = 1/3 W + 1/6 W  and  Req = 2W.

    The total resistance of the circuit is (2 + 4) W = 6 W  and
    I = 12 V/6 W = 2 A.

  2. 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) W  and
    I = 12/7 A.

  3. 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) W   and
    I = 12V/10 W = 1.2 A.

7.


In the Fig. for #7, the 10 W resistance is wired in parallel with the 15 W resistance from c to b’. This parallel grouping is then wired in series with 4.0 W resistor between a’ and b’. Finally this grouping is wired in parallel with another 10 W resistor.

1/Rcb’ = 1/10 W + 1/15 W = (3 +2)/30 W.
Rcb’ = 6 W.
Rab’ = (6 + 4) W = 10 W.
1/Rab = 1/10 W + 1/10 W = 2/10 W.
Rab = 5.0 W


8.



For the circuit in Fig. for #8 above,   I= e/(r + R).

For  I = 6.0 A and R = 2.0 W,
6.0 A = e/(r + 2.0 W)       (Equation 1)
For  I = 3.0 A and R = 7.0 W,
3.0 A = e/(r + 7.0 W)      (Equation 2)
From Eq. 1:  e = 6.0A r + 12 V.  From Eq. 2:  e = 3.0A r + 21V.
Thus, 6.0A r + 12 V = 3.0A r + 21 V.   r = 9.0V/3.0 A = 3.0 W.
Substituting r back into Eq. 1,  6.0 A = e/(3.0 + 2.0) W.   e = 30 V.

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) W- 6 V - I2(3 W) = 0   or
30 A = 4I1 + 3I2       (Equation 2)
For loop II,
6 V + (3 W)I2 - 4 V - (2 W)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     or    I2 = 2A.

Then from Eq. 2:               30A = 4I1 + 3(2A)   and   I1 = 6 A.

From Eq. 1,   I3 = 6A - 2A = 4A.

10.



Given q(t) = Qi e-t/RC.  When t = RC,  q (RC) = Qi e -RC/RC = Qi e-1 = Qi/e.
The time constant for the circuit is RC.
  1. 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)MW= 7.0 MW = 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.
  2. 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.

11.



  1. Vab = Vac + Vcb
    e = q/C + RI = q/C + Rdq/dt.
    q - eC = - RC dq/dt,  or separating variables
    dq/(q - eC) = - dt/RC.

    Integrating,
    Qo ∫q dq/(q - eC) = -1/RC o∫t dt
    ln (q - eC)/(Qo - eC) = - t/RC.
    Taking the antilog of both sides of the equation and rearranging,
    q(t) = Qo e-t/RC + eC(1 - e-t/RC ).
    I(t) = dq/dt = - Qo/RC e-t/RC + e/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 + e/R = (-100 + 40)V/500 W = 0.12 A.

  2. q(∞) = eC = 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 = e/(r + R), and the power delivered to the load resistor R, is P = I2R = e2R/(r + R)2.

For maximum power,
dP/dR = e2[(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 = e2R/(r + R)2.  For R = 0,  P = 0.  As R approaches a very large number, P approaches e/R which goes to zero.  In the Fig. for #12 b above, I have taken e = 2.0 V  and  r = 0.5 W.  Notice that P is a maximum at R = r = 0.5 W.

13.

  1. 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 W/2) = 3000 W  (Fig. 8d below).

    I1 = e/Rab = 6 V/3000 W = 2 x 10-3 A.
    Vac = I1(R/2) = 2 x 10-3 A(1000 W) = 2 V.
    I2 = I3 = Vac/R = 2 V/2000 W = 1 x 10-3 A.

    Notice that  I1 = I2 + I3.



  2. 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 = e/2R = 6 V/2000 W = 3 x 10-3A.







  3. From conservation of charge,
    I1 = I2 + I3    (Equation 1)
    From conservation of energy,
    Loop I:     e - I1R – I2R = 0            (Equation 2)
    Loop II:   I2R – I3R - q/C = 0             (Equation 3)
    Substituting Eq. 1 into Eq. 2:
    e - (I2 + I3)R – I2R = 0    (Equation 4)
    Solving for I2 in Eq. 4:
    I2 = (e - I3R)/2R               (Equation 5)
    Substituting Eq. 5 into Eq. 3:
    (e - I3R)/2 – I3R + q/C = 0      (Equation 6)
    Rearranging Eq. 6:
    e = 3I3R + 2q/C       (Equation 7)
    Recognizing that I3 = dq/dt,  Eq. 7 becomes :
    e = 3(dq/dt)R + 2q/C     (Equation8)
    The equation we solved for in the Outline for Circuits was
    e = (dq/dt)R + q/C      (Equation 9)
    and had solution q(t) = Ce(1 – e-t/RC) and time constant t = RC.

    Comparing Eq. 8 and Eq. 9, we see for Eq. 8 that e becomes e/2
    and the time constant t = 3RC/2.

    For this case,
    q(t) = (Ce/2)(1 – e-2t/3RC)  and
    I3(t) = dq/dt = (e/3R) e-2t/3RC      (Equation 10)
    I3(0) = (e/3R) e0 = (e/3R)   and
    I3(∞) = 0,  as in Part (b) above.
    From Eq. 5,
    I2 = (e - I3R)/2R = e/2R - I3/2 = e/2R - (e/6R) e-2t/3RC
    I2(t) = e/6R(3 - e-2t/3RC)          (Equation 11)
    I2(0) = e/3R  and
    I2(∞) = e/2R,  as in Part (b) above.
    From Eq. (1),
    I1 = I2 + I3 = (e/6R)(3 - e-2t/3RC) + (e/3R) e-2t/3RC
    I1(t) = (e/6R) (3 + e-2t/3RC)       (Equation 12)
    I1(0) = (2e/3R)  and
    I1(∞) = (e/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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Susan D. Kunk
Phyllis J. Fleming
August 8, 2002
April 30, 2003