Solution Set - Circuits

Phyllis Fleming Physics

Physics 108

Answers - Circuits

1.	The average current density is J = I/A = 20A/ π(1.63 x 10^-3 m)² = 2.4 x 10⁶ A/m². v_d = J/en = 2.4 x 10⁶ A/m²/(1.6 x 10^-19C)(8.47 x 10²⁸m^-3) = 1.8 x 10^-4m/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^-6m²)/1.56 x 10^-8Ω - m = 0.92 m.

2.	If the current I₁= 3.0 A in the 4.0 Ω resistor, V_cb = I₁(4.0 Ω) = 3.0 A(4.0 Ω) = 12 V. I₂= V_cb/6.0 Ω= 12 V/6.0 Ω = 2.0 A. I = I₁+ I₂= 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.

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”.
I₁= V_c’e’/R_c’e’ = 12 V/ 8 Ω = 3/2 A.
I₂ = V_c”e”/R_c”e”= 12 V/24 Ω = 1/2 A. Note that I₁ + I₂ = I.

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’) V_ab = (V_ab/R₁)(V_ab) = V_ab²/R₁.
P₂ = (I₂’) V_ab = (V_ab/R₂)(V_ab) = V_ab²/R₂.

Since R₁ > R₂, now P₂ > P₁.

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.

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.

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

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.

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) = Q_i e^-t/RC. When t = RC, q (RC) = Q_i e ^-RC/RC = Q_ie^-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^-6C/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 10⁶V/A.
The time constant =
RC = 7.0 x 10⁶V/A x 20 x 10^-6C/V = 140 C/A = 140 s.
q(t) = Q_ie^-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∫^qdq/(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^-6C/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^-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/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).

I₁= ε/R_ab = 6 V/3000 Ω = 2 x 10^-3A.
V_ac = I₁(R/2) = 2 x 10^-3 A(1000 Ω) = 2 V.
I₂ = I₃= V_ac/R = 2 V/2000 Ω = 1 x 10^-3A.

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/3RCI₂(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 I₃as a function of time is shown in Fig. 8g below.

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Website Designed By: Questions, Comments To: Date Created: Date Last Updated:		Susan D. Kunk Phyllis J. Fleming August 8, 2002 April 30, 2003