Question

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


A circuit consists of a 12V battery connected across a single resistor. If the current in the circuit is 3 A, calculate the size of the resistor. (R=4Ω)


If a clock expends 2 W of power from a 1.5 V battery, what amount of current is supplying the clock? ( I = 1.33A)


Tommy runs his juicer every morning. The juicer uses 90 W of Power and the current supplied is 4.5 A. How many volts are necessary to run the juicer? (V=20V)


A DC electric motor transforms 1.50 kW of electrical power into mechanical form. If the motor’s operating voltage is 300 volts, how much current does it “draw” when operating at full power output? (don’t forget to change SI prefixes,I=5A)



I need help in each one please

Answer 1

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

Answer 2

a) I=0.1 A

b) R=4 ohms

c) I=1.33 A

d) V=20V

e) I=5A

Solving for each part one by one:

Ohm's law

a) From ohm's law, it states that voltage is directly proportional to current and resistance.

$V = IR$

where

V is the supply voltage

R is the resistance

I is the current

Given:

Resistance, R = 200ohms

Voltage, V = 20V

To find:

Current, I=?

On substituting the values in the above formula:

$I =\frac{V}{R}\\\\I =\frac{20}{200}\\\\I = 0.1Amperes$

b) Using the ohm's law formula

V= IR

Given:

Voltage, V = 12volts

Current, I = 3A

To find:

Resistance, R=?

$R = \frac{V}{I}\\\\R = \frac{12}{3}\\\\R = 4ohms$

c) Power generated by the battery is expressed as:

P = IV

I = P/V

Given:

Power, P = 2Watts

Voltage, V = 1.5volts

$I = \frac{2}{1.5}\\\\I = 1.33A$

d) similarly, Power = current I * voltage V

V = P/I

Given:

P = 90watts

I = 4.5A

$V = \frac{90}{4.5} \\\\V = 20volts$

e)

Given:

Power, P= 1.5kW = 1500watts

Voltage, V = 300volts

$I = \frac{P}{V}\\\\I = \frac{1500}{300}\\\\I = 5A$

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