I = V/R, so 9V and 0.5 ohms is 18A (not volts as you wrote).
No limits to the equation except that wires overheat or even vapourize when there is too much current for that size wire.
A reason the formula might appear to fail is that the voltage source (battery) used may not supply enough current. It is as if thgere is an internal resistor in series with the supply terminals. A small 9V battery has like 10 ohms internal resistance.Thus if the internal resistor is 10 ohms, and we apply a short circuit (close to zero ohms) the current is still according to that 10 ohms, almost 1 Amp.
Yet another cause of confusion comes from ammeters (DMM on current ranges). These also have an internal resistance.
So for practical reasons, if you are doing tests with a small 9V battery, use resistors in the range of 1000 ohms so the current is of the order 10mA and all will be well.
No. Ohms Law, (simplified) is:
P = Power in Watts
E = Voltage in Volts,
I = Current in Amperes,
R = Resistance in Ohms
(Multiply goin\g up, divide going down)
Therefore,
P = 162
E = 9
I = 18
R = 0.5
9 / 0.5 = 18 AMPS, 18 X 9 = 162 Watts.
In the REAL world, this will not happen, however, because there is several ohms of resistance in the battery, so the actual current - and so total power dissipated will be considerably less. The situation would be quite different is you were using a heavy-duty regulated power supply instead of a battery, however these exact figures would never be attained or even closely approached in the real world except in a laboratory using VERY expensive equipment..
To answer4 your second question, Ohms Law ALWAYS rules! It is in effect for any and all circumstances where there is a voltage differential and current flows.
Nope. You would have 18 AMPS
Volts /div by/ Resistance= Amperage
That is what the formula is talking about.
You seem to be misunderstanding ohms as "percentage of Circuit impedance, where 1= 100%"... which is simply not the case. Voltage is the "pressure" of electricity in a circuit, Ohms is the resistance to that pressure, therefore Amperage is the flow rate of the voltage through that resistance, and can be computed simply by Ohms Law.
9 volts across .5 ohms give 18 amps, not 18 volts. If a resistor is linear, then V=I*R for all values.
Ohms law applies of course.
assuming load resistance is fixed, as the resistance of wire increases, entire circuit resistance increases
R=Rload+2*Rwire
and as R increases current in circuit is smaller.
you would not get 18V. Load would see lower voltage because of voltage drop on wires.
suppose your load is something with resistance of 50 Ohm and you have 9V battery and pair of wires, each with resistance of 0.001 Ohm
R=50+0.001+0.001=50.002
I=V/R=9/50.002=0.1799928 Amp
Load would see voltage reduced by voltage drop on wires.
each wire would drop
Vwire=I*Rwire=0.0001799928 V
so Load would only get
Vload= Vbatt - 2*Vwire = 9-2*0.0001799928 = 8.99964 v
but suppose you decided to replace wires with ones that are Rwire=0.5 Ohm. then we get
R=Rload+2*Rwire=50+0.5+0.5=51 Ohm
I=Vbatt/R=0.17647 Amp
and each wire would cause voltage drop of
Vwire=I*Rwire=0.0882353 V
so Load would get
Vload=Vbatt - 2*Vwire = 8.82353 V
if you somehow find wire that is even greater resistance, such as 1 Ohm each
load would see voltage of only 8.653846 V.
you can calculate power lost in wires
Pwire=I^2*R
but since we have two wires in circuit you need to double it
Ploss=2*R*I^2
when Rwire=0.001 Ohm,
Ploss=0.0000622833218W = 0.0622833218mW
when Rwire=0.5 Ohm
Ploss=0.0311416609W = 31.1416609mW
when Rwire=1Ohm
Ploss=0.0599112426W=59.9112426mW
V= IR is always applicable
Since wires have very little resistance (like .0001 Ohms in one foot of copper wire) does the V/R formula apply still? For example, if I have 9 volts at .5 Ohms, would I then have 18 volts? Or does the accuracy of the formula stop a 1?
