HELP ASAPPPPPPP !!!!

A 20-volt relay has a coil resistance of 200 ohms. How much current does it draw? (I=0.1A)

TEA [102]

Answer:

Explanation:

a) According to ohm's law

V = IR

V is the supply voltage

R is the resistance

I is the current

Given Resistance = 200ohms

Voltage = 20V

I = V/R

I = 20/200

I = 0.1Amperes

b) Using the ohm's law formula

V= IR

Where voltage = 12volts

Current I = 3A

Resistance R = V/I

R = 12/3

R = 4ohms

c) Power generated by the battery is expressed as P = IV

I = P/V

Given Power = 2Watts

V = 1.5volts

I = 2/1.5

I = 1.33A

d) similarly, power = current I × voltage V

V = P/I

Given P = 90watts

I = 4.5A

V = 90/4.5

V = 20volts

e) Given power = 1.5kW = 1500watts

Voltage = 300volts

I = P/V

I = 1500/300

I = 5A

8 0

3 years ago

Solving a series circuit, did I do this correctly?

nirvana33 [79]

The total resistance in the circuit is 16 Ohms.
The total current in the circuit is 0.5 Ampere.
The current at $R_1$ is 0.5 Ampere.
The current at $R_3$ is 0.5 Ampere.
The voltage drop at $R_1$ is 4 volts.
The voltage drop at $R_2$ is 2.5 volts.
The voltage drop at $R_3$ is 1.5 volts.
The total power consumed by the circuit is 4watts
The power consumed at $R_1$ is 2 watts
The power consumed at $R_2$ is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

$R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega$

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

The total resistance in the circuit = $R_{eq}$

$R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega$

The total current in the circuit = I

$V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A$ (Ohm's law)

For series combinations, the current in each resistor remains the same.

So, the current in $R_1, R_2 \&R_3$ :

$I_1= I_2= I_3=I=0.5 A\\$

The voltage drop across at $R_1$ = $V_1$

The current across $R_1$ = I = 0.5 A

$V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V$

The voltage drop across at $R_2$ = $V_2$

The current across $R_2$ = I = 0.5 A

$V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V$

The voltage drop across at $R_3$ = $V_3$

The current across $R_3$ = I = 0.5 A

$V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V$

The total power consumed by circuit:

$P= V\times I \\\\= 0.5 A\times 8 V = 4 watt$

Power consumed at $R_1$ :

$P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt$

Power consumed at $R_2$ :

$P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt$

Power consumed at $R_3$ :

$P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt$

Learn more about, current, voltage, resistance, and power of the circuit here:

brainly.com/question/11683246?referrer=searchResults

brainly.com/question/1430450?referrer=searchResults

5 0

2 years ago

PLZ HELP I DON'T UNDERSTAND!! a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, t

algol13

Explanation:

The mass of a ball, m = 2 kg

It is traveling with a speed of 10 m/s

The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2\times (10)^2\\\\=100\ J$

The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.

Hence, the required kinetic and potential energy is 100 J.

6 0

3 years ago

Fre.e points because you deserve it

bagirrra123 [75]

Answer:

thank you so much.. i appreciate your kindness

5 0

3 years ago

Read 2 more answers

H20 (water) is an example of a __, as it is made up of two elements.

vova2212 [387]

Answer:

Compound.

Explanation:

A compound is a substance formed when two or more elements are chemically joined. Water, salt, and sugar are examples of compounds. When the elements are joined, the atoms lose their individual properties and have different properties from the elements they are composed of.

8 0

4 years ago