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Chapter 1 – Electric Circuit Variables
Exercises
Exercise 1.2-1 Find the charge that has entered an element by time t when i = 8 t^2 – 4 t A, t ≥ 0. Assume q ( t ) = 0 for t < 0.
Answer:^3
( ) 2 C
q t = t − t
Solution:
2 3 2 3 2 0 0 0
8 4 A
( ) (0) (8 4 ) 0 2 2 C
t t t
i t t t
q t i d τ q τ τ d τ τ τ t t
= (^) ∫ + = (^) ∫ − + = − = −
Exercise 1.2-2 The total charge that has entered a circuit element is q ( t ) = 4 sin 3 t C when t ≥ 0 and q ( t ) = 0 when t < 0. Determine the current in this circuit element for t > 0.
Answer: ( ) 4sin 3 12 cos 3 A
d i t t t dt
Solution: ( ) 4sin 3 12 cos 3 A
dq d i t t t dt dt
Exercise 1.3-1 Which of the three currents, i 1 = 45 μ A, i 2 = 0.03 mA, and i 3 = 25 × 10–4^ A, is largest? Answer: i 3 is largest.
Solution:
i 1 = 45 μA = 45 × 10 -6^ A < i 2 = 0.03 mA = .03 × 10 -3^ A = 3 × 10 -5^ A < i 3 = 25 × 10 -4^ A
Exercise 1.5-1 Figure E 1.5-1 shows four circuit elements identified by the letters A , B , C , and D. (a) Which of the devices supply 12 W? (b) Which of the devices absorb 12 W? (c) What is the value of the power received by device B? (d) What is the value of the power delivered by device B? (e) What is the value of the power delivered by device D? Answers: (a) B and C , (b) A and D , (c)–12 W, (c) 12 W, (e)–12 W
(A)
3 A
(B)
6 A (C)
2 A
(D)
4 A
Figure E 1.5-
Solution: (a) B and C. The element voltage and current do not adhere to the passive convention in B and C so the product of the element voltage and current is the power supplied by these elements.
(b) A and D. The element voltage and current adhere to the passive convention in A and D so the product of the element voltage and current is the power delivered to, or absorbed by these elements.
(c) −12 W. The element voltage and current do not adhere to the passive convention in B , so the product of the element voltage and current is the power received by this element: (2 V)(6 A) = −12 W. The power supplied by the element is the negative of the power delivered to the element, 12 W.
(d) 12 W
(e) –12 W. The element voltage and current adhere to the passive convention in D , so the product of the element voltage and current is the power received by this element: (3 V)(4 A) = 12 W. The power supplied by the element is the negative of the power received to the element, −12 W.
P 1.2-4 The current in a circuit element is
t t i t t t
⎧^ <
⎪−^ <^ <
where the units of current are A
and the units of time are s. Determine the total charge that has entered a circuit element for t ≥ 0.
Answer: 0 2 2 4 2 4 ( ) where the units of 8 4 8 0 8
⎧^ <
⎪ −^ <^ <
t t t q t t t t
charge are C.
Solution:
t t
q t i τ d τ d τ
−∞ −∞ = (^) ∫ = (^) ∫ =0 C for t ≤ 2 so q (2) = 0.
t t (^) t q t = (^) ∫ i τ d τ+ q = (^) ∫ d τ = τ = 2 t −4 C for 2 ≤ t ≤ 4. In particular, q (4) = 4 C.
t t (^) t q t = (^) ∫ i τ d τ+ q = (^) ∫ − d τ+ = −τ + = 8 − t C for 4 ≤ t ≤ 8. In particular, q (8) = 0 C.
( ) 8 ( ) ( )^8 80
t t q t = (^) ∫ i τ d τ+ q = (^) ∫ d τ+ =0 C for 8 ≤ t.
P 1.2-5 The total charge q ( t ), in coulombs, that enters the terminal of an element is
2 ( 2)
3 t t 2
t q t t t e −^ − t
Find the current i ( t ) and sketch its waveform for t ≥ 0.
Solution:
(^2) ( (^2) )
2 t 2
t dq t i t i t t dt e −^ − t
⎩−^ >
P 1.2-6 An electroplating bath, as shown in Figure P 1.2-6, is used to plate silver uniformly onto objects such as kitchen ware and plates. A current of 600 A flows for 20 minutes, and each coulomb transports 1.118 mg of silver. What is the weight of silver deposited in grams?
Silver bar Object to be plated
Bath
i i
Figure P 1.2-
Solution:
5
C
= 450 A = 450
s C s mg Silver deposited = 450 20 min 60 1.118 = 6.0372 10 mg=603.72 g s min C
i
× × × ×
Section 1-3 Systems of Units
P 1.3-1 A constant current of 3.2 μA flows through an element. What is the charge that has passed through the element in the first millisecond? Answer: 3.2 nC
Solution:
Δ q = i Δ t = ( 3.2 × 10 −^6 A )( 1 × 10 −^3 s ) = 3.2 × 10 −^9 As = 3.2 × 10 −^9 C = 3.2 × 10 −^9 nC
P 1.3-2 A charge of 45 nC passes through a circuit element during a particular interval of time that is 5 ms in duration. Determine the average current in this circuit element during that interval of time. Answer: i = 9 μ A
Solution: 9 6 3
q i t
− − −
Δ ×
= = = ×
Δ ×
= 9 μA
P 1.3-3 Ten billion electrons per second pass through a particular circuit element. What is the average current in that circuit element? Answer: i = 1.602 nA
Solution
19 9 19
10 19
9
electron C electron C = 10 billion 1.602 10 = 10 10 1.602 10 s electron s electron electron C = 10 1.602 10 s electron C 1.602 10 1.602 nA s
i −^ −
−
−
⎡ ⎤ ⎡ × ⎤ ⎡ × ⎤ ⎡ × ⎤
× ×
= × =
P1.3-4 The charge flowing in a wire is plotted in Figure P1.3-4. Sketch the corresponding current.
Figure P1.3-
P1.3-
9 3 6 9 3 6 15 10 7.5 10 7.5 mA when 0 2 2 10 15 10 the slope of the versus plot 5 10 5 mA when 4 7 3 10 0 otherwise
t s
d i t q t q t s t s dt
− − − − − − ⎧ (^) × ⎪ =^ ×^ =^ <^ < ⎪ × ⎪ × = = = (^) ⎨− = − × = − < < ⎪ × ⎪ ⎪ ⎩
P1.3-6 The current in a circuit element is plotted in Figure P1.3-6. Determine the total charge that flows through the circuit element between 300 and 1200 μs.
Figure P1.3-
Solution:
1000 s 300 s q t i d "area under the curve between 300 s and 1000 s"
μ μ
( ) 9 ( 6 ) ( 9 )( 6 ) ( )^12
10 100 10 720 10 600 10 54 432 10 486 pC 2
q t −^ −^ −^ −^ −
= ⎜ × ⎟ × + × × = + × =
Section 1-5 Power and Energy
P1.5-1 Figure P1.5-1 shows four circuit elements identified by the letters A , B , C , and D. ( a ) Which of the devices supply 30 mW? ( b ) Which of the devices absorb 0.03 W? ( c ) What is the value of the power received by device B? ( d ) What is the value of the power delivered by device B? ( e ) What is the value of the power delivered by device C?
Figure P1.5-
Solution:
(a) A and D. The element voltage and current do not adhere to the passive convention in Figures P1.5- A and D so the product of the element voltage and current is the power supplied by these elements.
(b) B and C. The element voltage and current adhere to the passive convention in Figures P1.5- B and C so the product of the element voltage and current is the power delivered to, or absorbed by these elements.
(c) 30 mW. The element voltage and current adhere to the passive convention in Figure P1.5-1 B , so the product of the element voltage and current is the power received by this element: (5 V)( mA) = 30 mW. The power supplied by the element is the negative of the power received to the element, −30 W.
(d) −30 mW
(e) –30 mW. The element voltage and current adhere to the passive convention in Figure P1.5- 1 C , so the product of the element voltage and current is the power received by this element: ( V)(6 mA) = 30 mW. The power supplied by the element is the negative of the power received to the element, −30 W.
P 1.5-4 The current through and voltage across an element vary with time as shown in Figure P 1.5-4. Sketch the power delivered to the element for t > 0. What is the total energy delivered to the element between t = 0 and t = 25 s? The element voltage and current adhere to the passive convention.
5
30
(^0 1015 25) t (s)
v (volts)
(a)
5
30
(^0 1015 25) t (s)
i (amp)
(b) Figure P 1.5-
Solution: 30 for 0 t 10 s: = 30 V and = 2 A 30(2 ) 60 W 15
≤ ≤ v i t = t ∴ P = t = t
for 10 15 s: 10 30 V 80 V 5 ( ) 5 80 and ( ) 2 A 2 5 80 10 160 W
t v t t b v b
v t t i t t P t t t t
for 15 t 25 s: 5 V and ( ) A 10 (25) 0 b = 75 ( ) 3 75 A 5 3 75 15 375 W
v i t t b
i i t t P t t
( ) (^ )
10 15 2 25 0 10 15 10 15 25 2 2 3 2 (^0 10 )
Energy 60 160 10 375 15
30 80 10 375 15 5833.3 J
P dt t dt t t dt t dt
t t t t t
P 1.5-5 An automobile battery is charged with a constant current of 2 A for five hours. The terminal voltage of the battery is v = 11 + 0.5 t V for t > 0, where t is in hours. (a) Find the energy delivered to the battery during the five hours. (b) If electric energy costs 15 cents/kWh, find the cost of charging the battery for five hours. Answer: (b) 1.84 cents
Solution: a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). 5 3600 ( ) 2 5 3600(^ ) (^0 ) 0 3
= 441 10 J 441 kJ
t w Pdt vi d d t
= = τ = ⎛⎜^ + ⎞⎟ τ= + τ
× =
∫ ∫ ∫
b.) Cost = 441kJ 1 hr^ 15¢ 1.84¢ 3600s kWhr
× × =
P 1.5-7 Find the power, p ( t ), supplied by the element shown in Figure P 1.5-6 when v ( t ) = 8 sin 3 t V and i ( t ) = 2 sin 3 t A.
Hint:
(sin )(sin ) (cos( ) cos( ) ) 2
at bt = a − b t − a + b t
Answer: p ( t ) = 8 – 8cos 6 t W
i
v
Figure P 1.5-
Solution:
p t ( ) = v t i t ( ) ( ) = ( 8sin 3 t )( 2sin 3 t ) = 8 cos 0( − cos 6 t )= 8 −8cos 6 t W
Here is a MATLAB program to plot p ( t ):
clear
t0=0; % initial time tf=2; % final time dt=0.02; % time increment t=t0:dt:tf; % time
v=8sin(3t); % device voltage i=2sin(3t); % device current
for k=1:length(t) p(k)=v(k)*i(k); % power end
plot(t,p) xlabel('time, s'); ylabel('power, W')
P 1.5-8 Find the power, p ( t ), supplied by the element shown in Figure P 1.5-6. The element voltage is represented as v ( t ) = 4(1– e –2 t )V when t ≥ 0 and v ( t ) = 0 when t < 0. The element current is represented as i ( t ) = 2 e –2 t A when t ≥ 0 and i ( t ) = 0 when t < 0.
Answer: p ( t ) = 8(1 – e –2 t ) e –2 t^ W
i
v
Figure P 1.5-
Solution:
p t ( ) = v t i t ( ) ( ) = 4 1( − e −^2 t^ ) × 2 e −^2 t^ = 8 1( − e −^2 t^ ) e −^2 t W
Here is a MATLAB program to plot p ( t ):
clear
t0=0; % initial time tf=2; % final time dt=0.02; % time increment t=t0:dt:tf; % time
v=4(1-exp(-2t)); % device voltage i=2exp(-2t); % device current
for k=1:length(t) p(k)=v(k)*i(k); % power end
plot(t,p) xlabel('time, s'); ylabel('power, W')
P1.5-10 Medical researchers studying hypertension often use a technique called “2D gel electrophoresis” to analyze the protein content of a tissue sample. An image of a typical “gel” is shown in Figure 1.5-10 a. The procedure for preparing the gel uses the electric circuit illustrated in Figure 1.5-10 b. The sample consists of a gel and a filter paper containing ionized proteins. A voltage source causes a large, constant voltage, 500 V, across the sample. The large, constant voltage moves the ionized proteins from the filter paper to the gel. The current in the sample is given by
i t ( ) = 2 + 30 e − at mA
where t is the time elapsed since the beginning of the procedure and the value of the constant a is 1
hr
a =
Determine the energy supplied by the voltage source when the gel preparation procedure lasts 3 hours.
500 V
sample
i ( t )
(b)
(a)
Devon Svoboda, Queen’s University
Figure 1.5-10 ( a ) An image of a gel and ( b ) the electric circuit used to preparation a gel.
Solution:
0 0 3 0 3 3 0 0
3 16.3 19.3 J
T T
t
t
energy w t p t d t v t i t d t
e d t
d t e d t
e
−
−
−
Section 1.7 How Can We Check…?
P 1.7-1 Conservation of energy requires that the sum of the power absorbed by all of the elements in a circuit be zero. Figure P 1.7-1 shows a circuit. All of the element voltages and currents are specified. Are these voltage and currents correct? Justify your answer.
+ + –
-
2 A
2 A 3 A 5 A
–5 A
4 V
- (^) 5 V +
3 V
+
-
–2 V
+ (^) 1 V – Figure P 1.7-
Hint: Calculate the power absorbed by each element. Add up all of these powers. If the sum is zero, conservation of energy is satisfied and the voltages and currents are probably correct. If the sum is not zero, the element voltages and currents cannot be correct.
Solution: Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are:
(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W = 0 W The element voltages and currents satisfy conservation of energy and may be correct.
