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Anna71 [15]
2 years ago
5

Consider two copper wires with circular cross-sections and equal lengths. one wire has 3 times the diameter of the other. How do

the resistances of these two wires compare?
Physics
1 answer:
Masteriza [31]2 years ago
5 0

The correct option is (C)  The longer wire has 3 times the resistance of the shorter wire.

Resistance is a measure of the opposition to current flow in an electrical circuit.

Why do circuits need resistance?

Using ohms (Ω) as the unit, resistance acts as a gauge to quantify how easily current will flow through a circuit. When resistance falls, current rises, and when resistance rises, current falls. In order to make sure that current flows in circuits at the proper rate, resistors are crucial.

How can resistance be determined in a circuit?

By applying Ohm's Law, you can get the total resistance if you know the total current and voltage through the entire circuit: R = V / I. A parallel circuit, for instance, has a voltage of 9 volts and a combined current of 3 amps. Total resistance (RT) is equal to 9 volts / 3 amps, or 3.

Learn more about the Resistance in circuit with the help of the given link:

brainly.com/question/1851488

#SPJ4

I understand that the question you are looking for is "Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistances of these two wires compare?

A) Both wires have the same resistance.

B) The longer wire has 1/3 the resistance of the shorter wire.

C) The longer wire has 3 times the resistance of the shorter wire.

D) The longer wire has 9 times times the resistance of the shorter wire.

E) The longer wire has 27 times times the resistance of the shorter wire."

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Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

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              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

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Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

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