3 mA are flowing through an 18 V circuit, how much resistance (in kΩ) is in that circuit?

Answers

Answer 1

Answer:

6kΩ

Explanation:

If we assume that the entire circuit's current is 3 mA, then we can compute the resistance within the circuit with the application of Ohm's law:

V = IR

V/I = R

R = V / I

R = 18V / 3 mA

R = 6kΩ

Hence, the resistance of the circuit is 6kΩ.

Cheers.


Related Questions

True or false Osha engineering controls are the last strategy of control an employer should use for job hazards

Answers

Answer:

false

Explanation:

You have a motor that runs at 1.5 amps from a 12 volt power source, how many watts of power is it using?

Answers

Answer:

18 Watts

Explanation:

For this problem, we simply need to understand the relationship of power to voltage and current.  This relationship is derived from Ohm's law:

Power = Voltage * Current

Given this equation, we can say the following to find the power consumption of the motor:

Power = 12volts * 1.5amps

Power = 18 Watts

Hence, the motor is consuming 18 Watts of power.

Cheers.

You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the circuit?

Answers

Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

A flexible rectangular area measures 2.5 m X 5.0 m in plan. It supports an external load of 150 kN/m^2. Determine the vertical stress increase due to the load at a depth of 6.25 m below the corner of the footing.

Answers

Answer:

19.0476 kN/m^2

Explanation:

Given data:

Dimension of rectangular area = 2.5m x 5.0m

external load = 150 KN/m^2

load depth = 6.25 m

calculate the vertical stress increase due to load at depth 6.25

we will use the approximate method which is

[tex]V_{s} = \frac{qBL}{(B +Z)(L+Z)}[/tex]  ------- (1)

q = 150 kN/m^2

B = 2.5 m

L = 5 m

Z = 6.25 m

substitute the given values into the equation (1) above

hence the Vs ( vertical stress ) = 19.0476 kN/m^2

Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa. Determine:

a. the air-fuel ratio
b. the temperature at which the water vapor in the products will start condensing.

Answers

Answer:

44.59°c

Explanation:

Given data :

Total pressure = 105 kpa

complete combustion

A) Determine air-fuel ratio

A-F = [tex]\frac{N_{air} }{N_{fuel} } = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }[/tex]

N = number of mole

m = molar mass

A-F = [tex]\frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}[/tex]  =  22.2 kg air/fuel

hence the ratio of Fuel-air = 1 : 22.2

B) Determine the temperature at which water vapor in the products start condensing

First we determine the partial  pressure of water vapor before using the steam table to determine the corresponding saturation temp

partial pressure of water vapor

Pv = [tex]\frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )[/tex]

N watervapor ( number of mole of water vapor ) = 3

N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol

Pro = 105

hence Pv = ( 3/33.53 ) * 105 =  9.39kPa

from the steam pressure table the corresponding saturation temperature to 9.39kPa =  44.59° c

Temperature at which condensing will start = 44.59°c

An equation showing the products of propylene with their mole numbers is attached below

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