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3 years ago

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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?

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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Now, putting the given values into the above formula as follows.

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Answer:

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From the question we are told that

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The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

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