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Dafna11 [192]
3 years ago
14

Which of the following would decrease the resistance to the flow of an electric current through a body?

Physics
2 answers:
seraphim [82]3 years ago
5 0

On page 37 it explains that If two conductors are composed of the same material and have the same cross-sectional area but different lengths, the electrical resistance of the longer conductor will be greater than that of the shorter conductor.

So the correct answer is A. shortening the conductor

Nookie1986 [14]3 years ago
3 0
R = ρ L/A. R= resistance, ρ= resistivity, L= length of the conductor. A = area of the conductor. Resistance is directly proportional to the length of the conductor. So if length of the conductor is decreased, resistance will also decrease. Hence A is the correct option
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An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The curren
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CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

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a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

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