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solong [7]
3 years ago
8

We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will n

ow dissipate the most heat?
Physics
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer:

Explanation:

The wire which is connected in series will dissipate the most heat.

The reason behind this fact is that, for the wire connected in series, there is only one path for the flow of electrons that's why electron face more opposition and thus more heat dissipation. However, in parallel connecting wires, alternate paths are available for the flow of electrons. That's why the electrons do not face that much opposition as in series case as the path and the flow is divided.

You can understand this concept by the example of traffic on the roads. If there is a single road, then you have to face the more opposition and you will have the less passage to drive through. However if there are parallel roads, then the traffic will got more space to move. It will become easier for them and opposition will be reduced. Hope you understand it :)

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
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Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

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The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

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distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

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The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
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