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const2013 [10]
4 years ago
11

you are working in a physics lab where you have made a simple circuit with a battery and bulb in which part of your circuit is t

he current flow maximum through the bulb filament or through the battery if you reverse the polarity would there be any difference in the intensity of the bulb​
Physics
1 answer:
LiRa [457]4 years ago
3 0

Answer:

The current moves in the terminal.

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A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
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The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

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Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = \frac{V}{R}

where R is the resistance

R = \frac{pL}{r^{2} \pi }

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = \frac{\pi(r)^{2}V  }{pL}

q = \frac{\pi(r^{2} )Vt }{pL}

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

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