 Phyllis Fleming Physics Physics 108
 1 The average current density is J = I/A = 20A/ π(1.63 x 10-3 m)2 = 2.4 x 106 A/m2. 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. E = ρJ = (1.56 x 10-8 V-m/A)(2.4 x 106 A/m2) = 0.037 V/m. R = ρL/A   or L = RA/ρ = 0.012 Ω (1.2 x 10-6 m2)/1.56 x 10-8 Ω - m = 0.92 m.
 2 If the current  I1 = 3.0 A in the 4.0 Ω resistor, Vcb = I1 (4.0 Ω) = 3.0 A(4.0 Ω) = 12 V. I2 = Vcb/6.0 Ω = 12 V/6.0 Ω = 2.0 A. I = I1 + I2 = 3.0 A + 2.0 A = 5.0 A. Vac = IRac = (5.0 A)(5.0 Ω) = 25 V. Vab = Vac + Vcb = 25 V + 12 V = 37 V.
 3 Rc’e’ = (2 + 6) Ω= 8 Ω. 3 Ω is in series with the 21 Ω.  Rc”e” = (3 + 21) Ω = 24 Ω. Rc’e’ is in parallel with Rc”e”. 1/Rce = (1/8 + 1/24)Ω-1 = (3/24 + 1/24) Ω-1.   Rce = 6 Ω. Rab = Rac + Rce + Reb = (2 + 6 + 4) Ω = 12 Ω. I = Vab/Rab = 24 V/12 Ω = 2 A. Vce = IRce = 2A(6 Ω) = 12 V = Vc’e’ = Vc”e”. I1 = Vc’e’/Rc’e’ = 12 V/ 8 Ω = 3/2 A. I2 = Vc”e”/Rc”e” = 12 V/24 Ω = 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 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 = I2(9.5 + 20)Ω = 1 A2 (29.5 V/A) = 29.5 W. Power dissipated in the internal resistance = I2(0.5 Ω) = (1 A2) (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/Req = 1/3 Ω + 1/6 Ω  and  Req = 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/Rcb’ = 1/10 Ω + 1/15 Ω = (3 +2)/30 Ω. Rcb’ = 6 Ω. Rab’ = (6 + 4) Ω = 10 Ω. 1/Rab = 1/10 Ω + 1/10 Ω = 2/10 Ω. Rab = 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, I1 = I2 + I3  or  I3 = I1 - I2    (Equation 1) For loop I, 36 V - I1(0.5 + 2 + 1.5) Ω- 6 V - I2(3 Ω) = 0   or 30 A = 4I1 + 3I2       (Equation 2) For loop II, 6 V + (3 Ω)I2 - 4 V - (2 Ω)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. 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)MΩ= 7.0 MΩ = 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. 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 Vab = Vac + Vcb 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)/(Qo - εC) = - t/RC. Taking the antilog of both sides of the equation and rearranging, q(t) = Qo e-t/RC + εC(1 - e-t/RC ). I(t) = dq/dt = - Qo/RC e-t/RC + ε/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 + ε/R = (-100 + 40)V/500 Ω = 0.12 A. q(∞) = εC = 40 V(4.0 x 10-6 C/V) = 1.6 x 10-4C. 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 = I2R = ε2R/(r + R)2.

For maximum power,
dP/dR = ε2[(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 = ε2R/(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/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 Ω/2) = 3000 Ω  (Fig. 8d below). I1 = ε/Rab = 6 V/3000 Ω = 2 x 10-3 A. Vac = I1(R/2) = 2 x 10-3 A(1000 Ω) = 2 V. I2 = I3 = Vac/R = 2 V/2000 Ω = 1 x 10-3 A. Notice that  I1 = I2 + I3. 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 = ε/2R = 6 V/2000 Ω = 3 x 10-3A.  From conservation of charge, I1 = I2 + I3    (Equation 1) From conservation of energy, Loop I:      ε - I1R – I2R = 0            (Equation 2) Loop II:   I2R – I3R - q/C = 0             (Equation 3) Substituting Eq. 1 into Eq. 2: ε - (I2 + I3)R – I2R = 0    (Equation 4) Solving for I2 in Eq. 4: I2 = ( ε - I3R)/2R               (Equation 5) Substituting Eq. 5 into Eq. 3: ( ε - I3R)/2 – I3R + q/C = 0      (Equation 6) Rearranging Eq. 6: ε = 3I3R + 2q/C       (Equation 7) Recognizing that I3 = 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 I3(t) = dq/dt = ( ε/3R) e-2t/3RC      (Equation 10) I3(0) = ( ε/3R) e0 = ( ε/3R)   and I3(∞) = 0,  as in Part (b) above. From Eq. 5, I2 = ( ε - I3R)/2R = ε/2R - I3/2 = ε/2R - ( ε/6R) e-2t/3RC I2(t) = ε/6R(3 - e-2t/3RC)          (Equation 11) I2(0) = ε/3R  and I2(∞) = ε/2R,  as in Part (b) above. From Eq. (1), I1 = I2 + I3 = ( ε/6R)(3 - e-2t/3RC) + ( ε/3R) e-2t/3RC I1(t) = ( ε/6R) (3 + e-2t/3RC)       (Equation 12) I1(0) = (2 ε/3R)  and I1(∞) = ( ε/2R), as in Part (b) above. A plot of  I1, I2, and I3 as a function of time is shown in Fig. 8g below. Homepage Sitemap
 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming August 8, 2002 April 30, 2003