Question

calculate power? where Current is 8 A , Volt is 9 v​

Answer 1

✠ Given Question :

• calculate power? where Current is 8 A , Volt is 9 volt

✠ Required Solution :

★ Values Given to us :

• ➸ Current is 8 A

• ➸ Volt is 9 volt

★ Formula Used Here :

• ➸ P = V × I

★ Where :

• ➸ P = Power

• ➸ V = potential difference

• ➸ I = Current

✠ Putting Values in Formula :

• ➸ P = 8 × 9 Watt

• ➸ P = 72 Watt

⠀⠀⠀⠀★ Power is 72 Watt ★

______________________________

Answer 2

Answer:

A 9V battery has approximately a stored energy of:

560mAh⋅9V13600sh11000m≈18144VAs≈18kJ

A joule is a watt-second, or a newton-meter. Under the most ideal conditions, with perfectly efficient machines everywhere, there is enough stored energy in a 9V battery to apply your specified 600N force over a distance of:

18kJ1NmJ1600N1000k=30m

Your proposed solenoid, requiring maybe 25A at 9V, consumes electrical power at the rate of:

25A⋅9V=225W

Applying your specified 600N force, and given that power, we can solve for the velocity your solenoid, were it 100% efficient, could provide:

225W1JWsNmJ1600N=0.375m/s

So you see, even if we can extract all the stored energy of the 9V battery with 100% efficiency, there isn't a whole ton of it. Knowing that your ideal solenoid is moving at 0.375m/s, and that the battery has enough energy for it to move 30m, the runtime is:

30m1s0.375m=80s

Or we could calculate it from the battery energy and solenoid power:

18000Ws11225W=80s

But maybe it's enough. The question is how to do it efficiently. Electrical power in a resistance is given by:

P=I2R

The internal resistance of a 9V battery is maybe 1.5Ω, when fresh. It goes up as the battery drains. Your solenoid is probably at least another 1Ω. So at 25A, your resistive losses alone would be:

(25A)2(1.5Ω+1Ω)=1562.5W

Compare this to the power used by the ideal solenoid considered above (225W) and you can see this is an absurdly inefficient system. Just dealing with the heat from these losses will be a challenge. Of course, you can't actually get this out of a 9V battery, because the voltage lost over its internal resistance at 25A is:

25A⋅1.5Ω=37.5V

...which is more than the 9V supplied by the battery.

Besides the battery, or the solenoid, transferring 225W of electrical power is a problem itself. Because power is the product of voltage and current (P=IE), to move a lot of power you can have high current, or high voltage. But, even wires have resistance, and since power lost to this resistance is proportional to the square of current, it's more practical to move high amounts of electrical power at high voltage than it is at high current. This is why the electric utility transmits power over long distances at very high voltage.

So, if you want to move 225W at 9V, you must keep the resistance very low, to avoid resistive losses from being very high. That means fat wire (including the wire in your solenoid, which accounts for most of the wire in the circuit) and batteries with low internal resistance. You can also trade current for voltage, or voltage for current, in the design of your solenoid, as supercat's answer describes.