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GalinKa [24]
3 years ago
11

What is the difference between a series circuit and a parallel circuit?

Engineering
2 answers:
serg [7]3 years ago
7 0

Answer:

In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component. ... In a series circuit, every device must function for the circuit to be complete. If one bulb burns out in a series circuit, the entire circuit is broken.

Explanation:

posledela3 years ago
3 0

Answer:

Here is ur answer

HAVE A NICE DAY

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Plane wall of material A with internal heat generation is insulated on one side and bounded by a second wall of material B, whic
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A convergentâdivergent nozzle has an exit area to throat area ratio of 4. It is supplied with air from a large reservoir in whic
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Answer:

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3 years ago
You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(
Vsevolod [243]

Answer:

Output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Explanation:

Given:

dc gain A = 23.6 dB

Input signal V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)

Now convert gain,

A = 10^{\frac{23.6}{20} } = 15.13

DC gain at frequency f = 0 is given by,

  A = \frac{V_{out} }{V_{in} }

V_{out} =AV_{in}

V_{out} = 15.13 \times   0.18 \cos (2\pi (57000)t +18.3)

At zero frequency above equation is written as,

V_{out} = 2.72 \times \cos 18.3

V_{out} = 2.72

Now we write output voltage as input voltage,

V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Therefore, output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

7 0
3 years ago
A soil specimen was tested to have a moisture content of 32%, a void ratio of 0.95, and a specific gravity of soil solids of 2.7
belka [17]

Answer:

a. 0.9263

b. 0.4872

c. 13.83kN/m^{3}

Explanation:

moisture content (ω) = 0.32

void ratio (e) = 0.95

specific gravity (G_{s}) = 2.75

the degree of satruation (S) = \frac{w . G_{s} }{e} =0.32×2.75/0.95 = 0.9263

b. porosity (n) = \frac{e}{e + 1} = 0.95/(0.95 + 1)= 0.4872

c. dry unit weight (γ_{d}) = \frac{G_{s} . V_{w} }{1 + e}

taking specific unit weight of water (V_{w})= 9.81kN/m^{3}

γ_{d} = 2.75 × 1000/(1 + 0.95) = 13.83kN/m^{3}

5 0
3 years ago
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