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2 years ago

What is the potential difference across a 15 Ω resistor that has a current of 3.0 A?

Physics

2 answers:

Zepler [3.9K]2 years ago

7 0

Answer:

45V

Explanation:

R=V/I

V=I/R

V=15ohms•3.0 A

V=45V

ehidna [41]2 years ago

5 0

Answer:

45 V

Explanation:

Use Ohm's law:

V = IR

V = (3.0 A) (15 Ω)

V = 45 V

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You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

3 years ago

An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq

wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / $\sqrt{(R)^{2} + (Xl -Xc)^{2} }$

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0

3 years ago

A 17 kg box sitting on a shelf has a potential energy of 350 J. How high is the shelf? Round your answer to the nearest whole nu

Free_Kalibri [48]

Potential energy is mass * gravity * height. (m*g*h).

350 = 17*9.8*h <--350 is its energy, 17kg is its mass, and 9.8 is gravity's acceleration on the object. We now just need to solve for h.

h = 350/(17 * 9.8) = 2.1 meters, which, when rounded to the nearest whole meter, is 2 meters.

The shelf is 2 meters high.

3 0

2 years ago

Read 2 more answers

A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff

nexus9112 [7]

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

4 0

2 years ago

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Otrada [13]

It would be 1. B 2. A 3. A

7 0

3 years ago