Ask question

Ask question

All categories

Anestetic [448]

3 years ago

10

A 1.5 V battery is connected to a 1,000 μF capacitor in series with a 150 Ω resistor. a. What is the maximum current that flows

through the resistor during charging? b. What is the maximum charge on the capacitor? c. How long does the capacitor take to reach a potential of 1.0V?

1 answer:

Elza [17]3 years ago

3 0

Answer:

$0.01\ \text{A}$

$0.0015\ \text{C}$

$0.0608\ \text{s}$

Explanation:

$V_0$ = Voltage = 1.5 V

$C$ = Capacitance = $1000\ \mu\text{F}$

$R$ = Resistance = $150\ \Omega$

Current is given by

$I=\dfrac{V_0}{R}\\\Rightarrow I=\dfrac{1.5}{150}\\\Rightarrow I=0.01\ \text{A}$

Current flowing in the resistor is $0.01\ \text{A}$ .

Charge is given by

$Q=CV\\\Rightarrow Q=1000\times 10^{-6}\times 1.5\\\Rightarrow Q=0.0015\ \text{C}$

The charge on the capacitor is $0.0015\ \text{C}$ .

Voltage is given by

$V=V_0e^{-\dfrac{t}{RC}}\\\Rightarrow t=-RC\ln\dfrac{V}{V_0}\\\Rightarrow t=-150\times 1000\times 10^{-6}\times\ln\dfrac{1}{1.5}\\\Rightarrow t=0.0608\ \text{s}$

Time taken to reach 1 V is $0.0608\ \text{s}$ .

You might be interested in

The instrument which is commonly used to measure the intensity of radioactivity is called a

tatyana61 [14]

The instrument is a Geiger counter and is used to measure radioactive level around people's bodies.

4 0

3 years ago

Read 2 more answers

The diagram below shows the relative positions of Earth and the Sun at a certain time of the year. Based on the diagram, which s

AlladinOne [14]

Answer:

summer

Explanation:

Notice the higher density of the rays of the sun hitting straight the latitudes below the equator.

7 0

3 years ago

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w

lilavasa [31]

Answer:

Mutual inductance, $M=2.28\times 10^{-5}\ H$

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

$M=\dfrac{\mu_oN_1N_2A}{2\pi r}$

(b) It is given that,

$N_1=550$

$N_2=290$

Radius, r = 10.6 cm = 0.106 m

Area of toroid, $A=0.76\ cm^2=7.6\times 10^{-5}\ m^2$

Mutual inductance, $M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}$

$M=0.0000228\ H$

or

$M=2.28\times 10^{-5}\ H$

So, the value of mutual inductance of the toroidal solenoid is $2.28\times 10^{-5}\ H$ . Hence, this is the required solution.

8 0

3 years ago

Help ASAP plsssss:///

kenny6666 [7]

Answer:

c

Explanation:

8 0

3 years ago

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

Norma-Jean [14]

Answer:

1)

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:

$R=R_1+R_2+...+R_n$

2)

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:

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

3)

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:

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

4)

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:

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

4 0

3 years ago