Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall
True or False
Answer:
false
Explanation:
An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section
Answer:
Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s
Explanation:
Given data :
Manometer reading ; p1 - p2 = 45 mm of water
Pressure at section ( I ) p1 = 100 kPa ( abs )
temperature ( T1 ) = 25°C
Pw ( density of water ) = 999 kg/m3
g = 9.81 m/s^2
next we apply Bernoulli equation at section 1 and section 2
p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex] ---------- ( 1 )
considering ideal gas equation
Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )
R ( constant ) = 287 NM/kg.k
T = 25 + 273.15 = 298.15 k
P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2
substitute values into equation ( 2 )
= 100 * 10^3 / (287 * 298.15)
= 1.17 kg/m^3
Also note ; p1 - p2 = PwgΔh ------- ( 3 )
finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )
[tex]\frac{PairV^{2} _{2} }{2}[/tex] = PwgΔh
[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex] = 753.86
[tex]v_{2}[/tex] = 27.5 m/s
trevor moves a magnetic toy train away from a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement?
Answer:a
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A battery with an f.e.m. of 12 V and negligible internal resistance is connected to a resistor of 545 How much energy is dissipated by the resistor in 65 s?
Answer:
When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then
R
1
in Figure 1(a) could be the resistance of the screwdriver’s shaft,
R
2
the resistance of its handle,
R
3
the person’s body resistance, and
R
4
the resistance of her shoes.
Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)
Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.
Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).
To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.
According to Ohm’s law, the voltage drop,
V
, across a resistor when a current flows through it is calculated using the equation
V
=
I
R
, where
I
equals the current in amps (A) and
R
is the resistance in ohms
(
Ω
)
. Another way to think of this is that
V
is the voltage necessary to make a current
I
flow through a resistance
R
.
So the voltage drop across
R
1
is
V
1
=
I
R
1
, that across
R
2
is
V
2
=
I
R
2
, and that across
R
3
is
V
3
=
I
R
3
. The sum of these voltages equals the voltage output of the source; that is,
V
=
V
1
+
V
2
+
V
3
.
This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation
P
E
=
q
V
, where
q
is the electric charge and
V
is the voltage. Thus the energy supplied by the source is
q
V
, while that dissipated by the resistors is
q
V
1
+
q
V
2
+
q
V
3
.
Explanation:
the importance of reading a circuit diagram to interpret a wiring diagram?
Answer:
The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.
Explanation:
How do you describe sound? (SELECT ALL THAT APPLY.) PLEASE HELP AND SELECT ALL THAT APPLY PLEASE!! A. Sound waves have to have a medium to travel through. B. The volume of a sound is known as amplitude. C. Loud sounds have high amplitude and vibrate with more energy than soft sounds. D. Sound waves are compression waves that cause energy transfer in air molecules.
Answer:
Sound waves are compression waves that cause energy transfer in air molecules
Sound waves have to have a medium to travel through
Loud sounds have high amplitude and vibrate with more energy than soft sounds
Explanation:
Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.
Sound is an example of a mechanical wave hence it requires a material medium for propagation.
The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.