Physics, 13.04.2020 04:14 karsynraine9419

1) draw a simple circuit with a voltage source and four resistors wired in series

2) draw a simple circuit with a voltage source and three resistors wired in parallel

3) draw a complex circuit with a voltage source with three total resistors. Two of the resistors should be drawn in parallel with each other, then in series with the third.

4) draw a complex circuit with a voltage source and six total resistors. You may set them up as you choose but be sure to include both series and parallel in your circuit.

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Answer from: Redeyestudio53

In this circuit (see attachment #1), we have:

- A voltage source: in this case, we choose a battery. A voltage source is a device producing an electromotive force (in a battery, this is done by means of a chemical reaction), which is responsible for "pushing" the electrons along the circuit and creating a current. The electromotive force (emf) of the battery is also called voltage, and it is indicated with the letter V.

- Four resistors: a resistor is a device which opposes to the flow of current. The property that describes by "how much" the resistor "opposes" to the flow of current is called "resistance", and it is indicated with the letter R.

- In this circuit, the 4 resistors are in series. Resistors are said to be in series when they are connected along the same branch of the circuit, so that the same current flow across each of them.

- For resistors in series, the equivalent resistance of the circuit is given by the sum of the individual resistances:

R=R_1+R_2+...+R_n

In this circuit (see attachment #2), we have:

- A voltage source: as before, we have chosen a battery, providing an electromotive force to the circuit

- Three resistors wired in parallel. Resistors are said to be connected in parallel when they are connected along different branches, but with their terminals connected to the same point, so that each of them has the same potential difference across it.

- For resistors in parallel, the equivalent resistance of the circuit is calculated using the formula:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}$

In this circuit (see attachment #3), we have:

- A voltage source (again, we have choosen a battery)

- Three resistors, of which:

-- 2 of them are connected in parallel with each other

-- the 3rd one it is in series with the first two

If we call R_1,R_2 the resistances of the first 2 resistors in parallel, their equivalent resistance is:

$\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}\\\rightarrow R_{12}=\frac{R_1 R_2}{R_1+R_2}$

Then, these two resistors are connected in series with resistor R_3 ; and so, the total resistance of this circuit will be:

$R=R_{12}+R_3=\frac{R_1R_2}{R_1+R_2}+R_3=\frac{R_1R_2+R_3(R_1+R_2)}{R_1+R_2}$

In this circuit (see attachment #4), we have:

- A voltage source (again, a battery)

- We have 6 resistors, which are arranged as follows:

-- Two branches each containing 3 resistors

-- The two branches are in parallel with each other

So, the total resistance of the two branches are:

$R_{123}=R_1+R_2+R_3$

$R_{456}=R_4+R_5+R_6$

And since the two branches are in parallel, their total resistance will be:

$\frac{1}{R}=\frac{1}{R_{123}}+\frac{1}{R_{456}}\\\rightarrow R=\frac{R_{123}R_{456}}{R_{123}+R_{456}}=\frac{(R_1+R_2+R_3)(R_4+R_5+R_6)}{R_1+R_2+R_3+R_4+R_5+R_6}$

1) draw a simple circuit with a voltage source and four resistors wired in series 2) draw a simple

Answer from: Quest

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

Answer from: Quest

The answer is 12, for the points

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