Chapter 1 - Electric Circuit Variables

Prévia do material em texto

Design Problems 
 
DP 1-1 
The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25) 
= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A 
Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust 
the estimates of the maximum voltage and current and a Grade A device otherwise. 
 
DP 1-2 
( ) ( ) ( )8 8 8 820 1 0.03 0.6 1 Wt t t tp t e e e e− − − −= − × = − 
 
Here is a MATLAB program to plot p(t): 
 
clear 
 
t0=0; % initial time 
tf=1; % final time 
dt=0.02; % time increment 
t=t0:dt:tf; % time 
 
v=20*(1-exp(-8*t)); % device voltage 
i=.030*exp(-8*t); % device current 
 
for k=1:length(t) 
 p(k)=v(k)*i(k); % power 
end 
 
plot(t,p) 
xlabel('time, s'); 
ylabel('power, W') 
 
Here is the plot: 
 
 
 
The circuit element must be able to absorb 0.15 W. 
1-9 
Problems 
Section 1-2 Electric Circuits and Current Flow 
 
P1.2-1 
( ) ( )5 54 1 20t tdi t e edt − −= − = A 
 
P1.2-2 
 ( ) ( ) ( ) ( )5 5
0 0 0 0
4 40 4 1 0 4 4 4
5 5
t t t t tq t i d q e d d e d t eτ ττ τ τ τ τ− −= + = − + = − = +∫ ∫ ∫ ∫ 5− − C 
 
 
P1.2-3 
( ) ( ) ( ) 00 0 4 40 4sin 3 0 cos3 cos3 C3 3
t t tq t i d q d tτ τ τ τ τ= + = + = − = − +∫ ∫ 43 
 
 
P1.2-4 
( ) ( ) 0t tq t i d dτ τ−∞ −∞= =∫ ∫ τ = 0 C for t ≤ 2 so q(2) = 0. 
( ) ( ) ( ) 22 22 2 2t t tq t i d q dτ τ τ τ= + = =∫ ∫ = 2t−4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C. 
( ) ( ) ( ) 44 44 1 4 4t t tq t i d q dτ τ τ τ= + = − + = − +∫ ∫ = 8−t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C. 
( ) ( ) ( )
8 8
8 0 0
t t
q t i d q dτ τ τ= + = +∫ ∫ = 0 C for 8 ≤ t . 
 
 
P1.2-5 
 
 
( ) ( )
( )2 2
0 0
 ( ) 2 0 2
2 2t
t
dq t
i t i t t
dt
e t− −
⎧ <⎪= = ⎨⎪− >⎩
 < <
 
 
 
P1.2-6 
 
5
C = 600 A = 600
s
C s mgSilver deposited = 600 20min 60 1.118 = 8.05 10 mg=805 g
s min C
i
× × × ×
 
 
1-2 
Section 1-3 Systems of Units 
 
P1.3-1 
( )( ) 4000 A 0.001 s 4 Cq i tΔ = Δ = = 
 
P1.3-2 
9
6
3
45 10 9 10
5 10
qi
t
−
−
−
Δ ×= = = ×Δ × = 9 μA 
 
P1.3-3 
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
electrons
C1.602 10 1.602 nA
s
i − −
−
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤× × ×⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
× ×
= × =
 
 
 
Section 1-5 Power and Energy 
 
P1.5-1 
 
 
a.) ( )( )( ) 4 = = 10A 2hrs 3600s/hr = 7.2 10 Cq i dt i t= Δ ×∫ 
b.) ( )( )110 V 10 A 1100 WP v i= = = 
c.) 0.06$Cost = 1.1kW 2hrs = 0.132 $
kWhr
× × 
 
P1.5-2 
 ( )( )
3
 = 6V 10 mA 0.06 W
200 W s 3.33 10 s
0.06 W
P
wt
P
=
Δ ⋅Δ = = = × 
 
 
1-3 
Section 1-3 Systems of Units 
 
P1.3-1 
( )( ) 4000 A 0.001 s 4 Cq i tΔ = Δ = = 
 
P1.3-2 
9
6
3
45 10 9 10
5 10
qi
t
−
−
−
Δ ×= = = ×Δ × = 9 μA 
 
P1.3-3 
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
electrons
C1.602 10 1.602 nA
s
i − −
−
−
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤× × ×⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
× ×
= × =
 
 
 
Section 1-5 Power and Energy 
 
P1.5-1 
 
 
a.) ( )( )( ) 4 = = 10A 2hrs 3600s/hr = 7.2 10 Cq i dt i t= Δ ×∫ 
b.) ( )( )110 V 10 A 1100 WP v i= = = 
c.) 0.06$Cost = 1.1kW 2hrs = 0.132 $
kWhr
× × 
 
P1.5-2 
 ( )( )
3
 = 6V 10 mA 0.06 W
200 W s 3.33 10 s
0.06 W
P
wt
P
=
Δ ⋅Δ = = = × 
 
 
1-3 
P1.5-3 
 30for 0 t 10 s: = 30 V and = 2 A 30(2 ) 60 W
15
v i t t P t≤ ≤ = ∴ = = t 
( ) ( )
( )( ) 2
25for 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
≤ ≤ = − + ⇒ = ⇒ =
= − + = ⇒ = − + = − +
 
( )( )
30for 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
i i t t
P t t
≤ ≤ = = − +
= ⇒ ⇒ = − +
b
∴ = − + = − +
 
 
 
 
 
( ) ( )10 15 2520 10 15
15 25102 2 3 2
0 10 15
Energy 60 160 10 375 15
10 1530 80 375 5833.3 J3 2
P dt t dt t t dt t dt
t t t t t
= = + − + −
= + − + − =
∫ ∫ ∫ ∫
 
 
 
1-4 
P1.5-4 
 a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). 
( ) ( )5 36005 3600 2
0 0
0
3
0.5 0.5 2 11 22
36003600
= 441 10 J 441 kJ
t
w Pdt vi d d tττ τ⎛ ⎞= = = + = +⎜ ⎟⎝ ⎠
× =
∫∫ ∫ τ 
 
 
b.) 1 hr 10¢Cost = 441kJ 1.23¢
3600s kWhr
× × = 
 
 
P1.5-5 
( ) ( ) ( ) ( ) ( )1 1 14cos3 sin 3 sin 0 sin 6 sin 6 W
12 6 6
p t v t i t t t t t⎛ ⎞= = = + =⎜ ⎟⎝ ⎠ 
( ) 10.5 sin 3 0.0235 W
6
p = = 
( ) 11 sin 6 0.0466 W
6
p = = − 
 
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*cos(3*t); % device voltage 
i=(1/12)*sin(3*t); % device current 
 
for k=1:length(t) 
 p(k)=v(k)*i(k); % power 
end 
 
plot(t,p) 
xlabel('time, s'); 
ylabel('power, W') 
1-5 
 
 
 
P1.5-6 
( ) ( ) ( ) ( )( ) ( )8sin 3 2sin 3 8 cos 0 cos 6 8 8cos 6 Wp t v t i t t t t t= = = − = − 
 
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=8*sin(3*t); % device voltage 
i=2*sin(3*t); % device current 
 
for k=1:length(t) 
 p(k)=v(k)*i(k); % power 
end 
 
plot(t,p) 
xlabel('time, s'); 
ylabel('power, W') 
1-6 
P1.5-7 
( ) ( ) ( ) ( ) ( )2 2 2 24 1 2 8 1 Wt t t tp t v t i t e e e e− − − −= = − × = − 
 
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(-2*t)); % device voltage 
i=2*exp(-2*t); % device current 
 
for k=1:length(t) 
 p(k)=v(k)*i(k); % power 
end 
 
plot(t,p) 
xlabel('time, s'); 
ylabel('power, W') 
 
 
P1.5-8 
 =3 0.2=0.6 W
 0.6 5 60=180 J
P V I
w P t
= ×
= Δ = × × 
 
 
1-7 
	Chapter 1 - Section 1
	Chapter 1 - Section 2
	Chapter 1 - Section 3
	Chapter 1 - Section 5