Answer:
The general rule of thumb is that the SMALLER a substance's atoms and the STRONGER the bonds, the harder the substance. Two of the strongest forms of chemical bonds are the ionic and covalent bonds.
Explanation:
Answer:
The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Explanation:
This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Therefore this is the formula for Half wave rectifier
Vrms = Vm/2 and Vdc
= Vm/π:
Where,
Vrms = rms value of input
Vdc = Average value of input
Vm = peak value of output
Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.
Answer:
V₂ = 20 V
Vt = 20 V
V₁ = 20 V
V₃ = 20 V
I₁ = 10 mA
I₃ = 3.33 mA
It = 18.33 mA
Rt = 1090.91 Ω
Pt = 0.367 W
P₁ = 0.2 W
P₂ = 0.1 W
P₃ = 0.067 W
Explanation:
Part of the picture is cut off. I assume there is a voltage source Vt there?
First, use Ohm's law to find V₂.
V = IR
V₂ = (0.005 A) (4000 Ω)
V₂ = 20 V
R₁ and R₃ are in parallel with R₂ and the voltage source Vt. That means V₁ = V₂ = V₃ = Vt.
V₁ = 20 V
V₃ = 20 V
Vt = 20 V
Now we can use Ohm's law again to find I₁ and I₃.
V = IR
I = V/R
I₁ = (20 V) / (2000 Ω)
I₁ = 0.01 A = 10 mA
I₃ = (20 V) / (6000 Ω)
I₃ = 0.00333 A = 3.33 mA
The current It passing through Vt is the sum of the currents in each branch.
It = I₁ + I₂ + I₃
It = 10 mA + 5 mA + 3.33 mA
It = 18.33 mA
The total resistance is the resistance of the parallel resistors:
1/Rt = 1/R₁ + 1/R₂ + 1/R₃
1/Rt = 1/2000 + 1/4000 + 1/6000
Rt = 1090.91 Ω
Finally, the power is simply each voltage times the corresponding current.
P = IV
Pt = (0.01833 A) (20 V)
Pt = 0.367 W
P₁ = (0.010 A) (20 V)
P₁ = 0.2 W
P₂ = (0.005 A) (20 V)
P₂ = 0.1 W
P₃ = (0.00333 A) (20 V)
P₃ = 0.067 W
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I had a couple of answers for this, but when I checked nothing
was right, so im not sure.
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