A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor
1 answer:
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Answer:
The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh
However, the average power P, required for a year, t = ? Kw
Expressing their relationship, we will have
U = P x t
Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)
t = 8760 hours
P =
P = 7134.7Kw
Hence, the average power required during the year is 7,135Kw
Now to calculate the energy used by the power plant in a year (in quads)?
Recall, Efficiency, η = Power Output/Power Input (100)
so, we have
η = P₀/P₁, given
0.45 =
P₁ = 15,855Kw
the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules
So to convert this to quads, Note;
1 quads of energy = 10¹⁵ Joules
The total energy used is 0.000000139 quads
Now to find the cubic feet of natural gas required to generate this power?
Note: 0.29Kwh of Power generated = 1 cubic feet of natural gas used
Since, the power plant generated = 62500000000Kwh
The cubic feet of natural gas used =
Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
