Ask question

Ask question

All categories

3 years ago

11

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

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

You might be interested in

An engineer is tasked to design a combinational circuit with three inputs x, y, z and one output F to satisfy the following cond

pashok25 [27]

A bc i know that is the answer

3 0

3 years ago

Plane wall of material A with internal heat generation is insulated on one side and bounded by a second wall of material B, whic

viktelen [127]

Sorry❤

Have a nice day ✨

8 0

3 years ago

A convergentâdivergent nozzle has an exit area to throat area ratio of 4. It is supplied with air from a large reservoir in whic

ryzh [129]

Answer:

Angle of discharge make at the edge of tube=64.9 degrees.

5 0

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