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A food processor draws 8.47 A of current when connected to a potential difference of 110 V.

a. What is the power consumed by this appliance?

b. How much electrical energy is consumed by this food processor monthly (30 days) if it is used on average of 10.0 min every day?

c. Assume that the price of electrical energy is 7.00 $/kWh. What is the monthly cost of using this food processor?

2. The electric meter in a house indicates that the refrigerator consumes 70.0 kWh in a week.

a. What is the power consumption of the refrigerator?

b. Assuming it is connected to a potential difference of 120 V , how much current does the refrigerator draw?

3. The heating element of an electric broiler dissipates 2.8 kW of power when connected to a potential difference of 120 V.

a. What is the resistance of the element?

b. How much current does the broiler draw? Use two ways to find out, and verify your answer.

Respuesta :

Answer:

i want the answer

Explanation:i want the answer

a. The power consumed by this appliance is 931.7W

b. The electrical energy is consumed by this food processor monthly (30 days) if it is used on average of 10.0 min every day is 4.65kWh

c. The  monthly cost of using this food processor is 32.55 $

What is power?

The rate of consumption of energy is power. it is given by the product of potential difference(V) and current(I)

a. P = VI  

here V  = 110V . I = 8.47 A

P = 110 x 8.47 = 931.7W

hence power consumed is 931.7 W

b. electrical energy  consumed by this food processor monthly (30 days) if it is used on average of 10.0 min every day is

E = P/t

Here, P = 931.7 W = 0.93kW

t = 10 x 30 = 300 min = 5 h

So, E =0.93 x 5 =  4.65 kWh

hence electrical energy consumed is 4.65kWh

c. The monthly cost of using the food processor is

given ,

Cost of one unit is 7.00$

Cost of 4.65 unit will be  7.0 x 4.65 =  32.55$

monthly cost of using the food processor  is  32.55$

2.

a.  Power consumption of refrigerator is  416 W

P = E/t  

Here, E = 70 kwh . t = 1 week = 168 h

So, P=  70/168 = 0.416 kW = 416 W

b. Current drawn by the refrigerator  is 3.46A

P = VI

Here, V = 120V, P = 416 W

So, I = P/V = 416/120 = 3.46 A

3.

a. The resistance of the element is 5.14 Ω

b. Current drawn by the boiler is 23,33A

What is resistance?

Resistance is the hindrance caused in the flow of current

a. The resistance is given by [tex]R = v^{2} / P[/tex]

here V = 120V , P = 2.8kW = 2800W

R = (120 x 120)/ 2800 = 5.14Ω

Hence Resistance of the boiler is 51.4Ω

b.   V = IR so, I = V /R

Here, V = 120V , R = 5.14Ω

So, I = 120/51.4 = 23.33 A

Hence current drawn by the boiler is 23.33A

The alternate way to find the current is by using formula

P = VI  

I = P/V

Since, P = 2800W,   V = 120V

So, I = 2800/120 =23.33 A

Hence Current comes out to be the same from both the method which verifies the answer .

To know more about power here

brainly.com/question/12584580

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