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mel-nik [20]
4 years ago
6

An electric car is being designed to have an average power output of 4,600 W for 2 h before needing to be recharged. (Assume the

re is no wasted energy.)
How much energy would be stored in the charged batteries? =J
The batteries operate on 80 V. What would the current be when they are operating at 4,600 W? =A
To be able to recharge the batteries in 1 h, how much power would have to be supplied to them? =W
Physics
1 answer:
MAVERICK [17]4 years ago
7 0

Answer:

a)3312 x 10⁴ J

b)I = 57.5 A

c)9200 W

Explanation:

Given that

P =4600 W

Time t= 2 h = 2 x 3600 s= 7200 s

We know that

1 W = 1 J/s

a)

Energy stored in the battery = P .t

                                              =4600 x 7200 J

                                            =3312 x 10⁴ J

b)

We know that power P given as

P = V .I

V=Voltage  ,I =Current

4600 = 80 x I

I = 57.5 A

c)

The energy supplied = 4600 x 2 = 9200 W

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A scientist examines a large pot of boiling water and a small cup of boiling water. The scientist determines that the large pot
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Answer:

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3 years ago
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Slav-nsk [51]

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3 years ago
Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so
mote1985 [20]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

4 0
3 years ago
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