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Problem 2.8
Solution:
Known quantities:
Recharging current and recharging voltage
Find:
a) Total transferred charge
b) Total transferred energy
Analysis:
∫
Q = area under the current vs . time curve = Idt
1
1
a) = (4)(0.5 x 60)(60) + 6(0.5 x 60)(60) + (2)(1.5 x 60)(60)
2
2
1
+ 4(1.5 x 60)(60) + (4)(60)(60 ) = 48,600 C
2
Q = 48,600 C
dw
b)
= p so w = ∫ pdt = ∫ vidt
dt
3
v=9+
t V, 0 ≤ t ≤ 10800 s
3 x 60 x 60
4
i1 = 10 t A, 0 ≤ t ≤ 1800 s
0.5 x 60 x 60
2
(t -1800) A, 1800 ≤ t ≤ 7200 s
i2 = 6 1.5 x 60 x 60
4
i 3 = 12 (t - 7200) A, 7200 ≤ t ≤ 10800 s
1.0 x 60 x 60
where i = i1 + i 2 + i3
Therefore,
w=
1800
∫ vi dt + ∫
0
1
7200
1800
vi2 dt +
10800
∫
7200
vi3 dt
Problem 2.14
Solution:
Known quantities:
Circuit shown in Figure P2.14.
Find:
The unknown currents.
Analysis:
Applying KCL at the node: i − 2 − 6 + 5 = 0 or − i + 2 + 6 − 5 = 0
thus i = 3 A which means that a 3-A current is leaving the node.
Problem 2.20
Solution:
Known quantities:
Circuit shown in Figure P2.20.
Find:
Determine power dissipated or supplied for each power
source.
HINT: Think about eference direction for voltages and
currents.
Analysis:
Element A:
P = vi = (-12V)(-25A) = 300W (dissipating)
Element B:
P = vi = (15V)(25A)= 375W (dissipating)
Element C:
P = vi = (27V)(-25A) = −675W (supplying)
Problem 2.23
Solution:
Known quantities:
Circuit shown in Figure P2.24.
Find:
Determine power absorbed or power delivered and
corresponding amount.
Analysis:
P = (100V)(- 4A) = − 400 W A supplies
P = (10V)(4A) = 40 W B absorbs
P = (100V)(-1A) = −100W C supplies
P = (− 10V)(− 1A) = 10W D absorbs
P = (90V)(5A ) = 450W E absorbs
Problem 2.37
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.36 with source voltage, v = 24V ; and resistances,
Ro = 8Ω,R1 = 10Ω,R2 = 2Ω .
Find:
a)
b)
c)
d)
e)
The equivalent resistance seen by the source
The current i
The power delivered by the source
The voltages v1 and v 2
The minimum power rating required for R1
Analysis:
Req = R0 + R1 + R2 = 8 + 10 + 2 = 20Ω
V
24 V
V − Req i = 0 , therefore i =
=
= 1.2A
Req 20Ω
a) The equivalent resistance seen by the source is
b) Applying KVL:
c) Psource = Vi = 24V ⋅ (− 1.2 A) = −28.8 W → Power is generated and supplied to the loads
d) Applying Ohm's law:
e)
v1 = R1i = 10Ω ⋅1.2A = 12 V , and v 2 = R2i = 2Ω ⋅1.2A = 2.4 V
P1 = R1i = 10Ω ⋅ (1.2A ) = 14.4 W , therefore the minimum power rating for R1 should be greater
2
2
than the generated power. Here 16 W can be suggested.
Problem 2.39
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.38.
Find:
The power delivered by the dependent source.
Analysis:
i=
24V
24
=
A=2A
(7 + 5)Ω 12
i source = 0.5i 2 = 0.5 ⋅ (4 ) = 2 A
The voltage across the dependent source (+ ref. taken at the top) can be found by KVL:
−v D + (2 A)(15Ω) + 24V = 0 ⇒ v D = 54 V
Therefore, the power
PD = v D isource = 54 ⋅ (− 2) = −108 W. which means that the power is generated by the dependent source.
Problem 2.49
Solution:
Known quantities:
Circuits of Figure 2.49.
Find:
Equivalent resistance.
Analysis:
(R 6 || R 4 ) + R 5 = (1,000 Ω || 100 Ω) + 9.1 Ω ≈ 100 Ω , resulting in
the circuit shown below.
Therefore, the equivalent resistance is
Req = (100||R3||R2 ) + R1 = (100||100||1000) + 5 = 52. 6 Ω
R1
R2
R
3
100 ?
Problem 2.55 is removed from the homework list.
Problem 2.66
Solution:
Known quantities:
Schematic of the circuit shown in Figure P2.66 with
resistances, R1 = 2.2kΩ, R2 = 18kΩ , R3 = 220kΩ, R4 = 3.3kΩ .
Find:
The equivalent resistance between A and B.
Analysis:
Shorting nodes C and D creates a single node to which all four resistors are connected.