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We know that $$P = I V$$ where $P$ is power, $V$ is voltage, and $I$ is current.

If voltage or current is increased, what happens to the power? For example, if $$V = 5 \, \text{Volts}, \quad I = 2 \, \text{Amps}$$ then we have $$P = 10 \, \text{Watts}.$$ Now, if the voltage is multiplied by $4$, how does that affect $I$ and $P$? If the current is multiplied by $4$, how does that affect $V$ and $P$?

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    $\begingroup$ How do you amplify the voltage? If you assume current is constant, it is obvious that power will increase. But in real applications the current may decrease. $\endgroup$ – jinawee Jan 20 '15 at 12:37
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Now, if the voltage is multiplied by 4, how does that affect $I$ and $P$?

Assuming your load is a resistor, then your original load resistance was 2.5Ω (i.e. 5V/2A). Therefore, if you increased the voltage to 20V, your resistance would remain the same (i.e. 2.5Ω), current would increase to 8A (20V/2.5Ω), and power consumption would increase to 160W (20V*8A).

If the current is multiplied by 4, how does that affect $V$ and $P$?

Assuming your load is a resistor and you do not change the original value of your resistor (i.e. 2.5Ω), then you would have to increase the voltage of the circuit to 20V to increase current in the circuit to 8A (20V/2.5Ω).

Alternatively, you could also create an 8A current flow within your circuit by keeping the same voltage (i.e. 5V) & changing your resistor value by mathematically solving for the resistor value that would be required (e.g.: 8A = 5V/?Ω ==> 5V/8A = 0.625Ω).

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Does their values changes if so what will be their estimated or resultant values.

INSUFFICIENT DATA FOR MEANINGFUL ANSWER

To answer this question, the relationship between the voltage and current variables given must be specified.

For example, if the voltage and current variables are the voltage across and current through a resistor, then we know

$$V = IR$$

and thus that the current will change proportionally with the voltage.

However, if a diode instead of a resistor, we have

$$I = I_S e^{\frac{V}{nV_T}}$$

and thus the current will change exponentially with the voltage.

In summary, if you specify the relationship between the voltage and current variables, you will probably be able to answer your own question.

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Power(P) = Voltage(V) * Current(I)

That law describes the relationship between power, voltage, and current in a conductor. It means that, if you measure the current flowing in the conductor, and you measure the voltage difference from one end of the conductor to the other at the same instant, then the product of voltage and current will be the rate at which heat is generated in the conductor at the same instant.

"Amplify" has no meaning in that situation.

Current can only flow in the conductor if some external circuit is forcing it. If the conditions at some other moment in time are different, then V, I, and P all will be different at that moment too, but they still will obey the same law.

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First of all power will alway be same i.e VI (primary side of amplifier) = VI (secondary side of amplifier). However V and I individually can change. So in your case:

Case 1: Voltage= 20 current will become I = 0.5 amp so net power = 10 watt

Case 2: If current I = 8 amps V will be = 10/8 Volts so net power = 10 watts

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  • $\begingroup$ Please note that using all caps denotes shouting on SE sites, and is never appreciated. $\endgroup$ – 299792458 Jul 13 '15 at 4:59
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The power will remain the same for a particular load as we are not changing the load. so if we increase the voltage, the current will decrease to make the net power consumed by the load same as before. If we increase the current, the voltage will decrease for making the power same. The power will only change when we changes the load.

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