Which of the following would decrease the resistance in a wire?

Physics

1 answer:

padilas [110]3 years ago

6 0

<u>Increase the thickness of the wire</u> would decrease the resistance in a wire

Explanation:

Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.

Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.

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Does Anybody Know The Answers?

ch4aika [34]

Answer:

I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry

Explanation:

$9^{2} + 12^2 = x^2\\81 + 144= x^2\\\sqrt{225} = \sqrt{x} \\ 15=x\\\\ 2.\\x^2+12^2+=13^2\\x^2+144 =169\\x^2 = 25\\\sqrt{x^2 =\sqrt{25\\\\$

x=5

$3.\\12^2+32^2 = x^2\\34.176= x$

5,12,13

$\frac{x}{4} ,\frac{12}{4} ,\frac{20}{4}\\\\\frac{x}{4},3,5 \\\\x=16\\\\12. \\x^2 + 48^2=50^2\\\\x^2=196\\x=14$

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5 0

3 years ago

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

$V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA$