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Vlad1618 [11]
3 years ago
11

A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 4.6 Ω resistor is conne

cted across the battery terminals, what is the terminal voltage and what is the current through the 4.6 Ω resistor?

Physics
2 answers:
rosijanka [135]3 years ago
5 0

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

san4es73 [151]3 years ago
4 0

Answer:

Terminal voltage = 8.36 V

Current = 1.82 A

Explanation:

E.M.F of battery = 12V

Internal resistance of battery (r) = 2Ω

Resistance of resistor (R) = 4.6Ω

Now the formula for terminal voltage across the battery is;

V = ε - Ir

Where ε is EMF and I is electric current

Using ohms law, we know that V = IR and I = V/R.

Thus, let's put V/R for current in the potential difference equation;

V = ε - r(V/R)

Thus, lets make V the subject of the formula ;

V + (rV/R) = ε

V(1 + r/R) = ε

So, V = ε/(1 + r/R)

V = 12/(1 + (2/4.6))

V = 12/(1 + 0.4348)

V = 12/1.4348 = 8.36 V

Thus from V=IR, we can find current. So 8.36 = I(4.6)

I = 8.36/4.6 = 1.82 A

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A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
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Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

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E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

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