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I have seen ohms law like this V=IR and I=V/R. V=IR tells me when you increase resistance Voltage increases. I=V/R tells me when you increase resistance Current decreases. So if I added a resistor to a circuit the voltage would increase and the current would decrease? It makes sense but I have also seen booster circuits. They always use inductors but if you just need a small voltage increase couldn't you just use the right resistor? If so why don't people do it?

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  • \$\begingroup\$ You're leaving out the energy part. You can't get money for nothing and energy for free. \$\endgroup\$ – Matt Young Nov 11 '14 at 13:21
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The V in Ohm's Law is very specifically the voltage drop across a resistor (or resistive element) given a current. The voltage will always drop in the direction of current flow. It can never increase in voltage through a resistor.

So, you said "when you increase resistance voltage increases". This is true, except it's not the voltage relative to ground that's increasing. It's the voltage drop across the resistor that's increasing. So the voltage relative to ground on the downstream side of the resistor goes lower.

Personally, I've always thought Ohm's Law should be presented as $$V_{drop} = IR$$ to avoid exactly this confusion.

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  • \$\begingroup\$ This answer needs to be clearer. The OP seems to confused about the direction current flow and how that appears to boost the voltage. Your \$ V_{drop}\$ contradicts your statement that "voltage will always drop ..." as it gives a positive result/value. \$\endgroup\$ – placeholder Aug 26 '14 at 9:41
  • \$\begingroup\$ Alright thanks everyone I totally misunderstood ohms law. I get it now though. \$\endgroup\$ – Trevor Aug 26 '14 at 12:41
  • \$\begingroup\$ @Placeholder, I'm not sure how I could be clearer, as I feel my answer addressed the exact concern you raised. The contradiction you pointed out is explained in the 2nd paragraph. You're entitled to your opinion of course, but I'm open to suggestions to reverse your downvote. \$\endgroup\$ – Dan Laks Aug 26 '14 at 14:06
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I assume you mean a booster converter or step-up converter when you refer to a "booster circuit". Without going into much detail, it works by rapidly switching the current in a circuit with a inductor and a diode in series. The current generated when the magnetic field around the inductor is collapsing causes the load to see a higher voltage. See the example here: http://en.wikipedia.org/wiki/Boost_converter

A resistor doesn't create (much) of an magnetic field so you will not get any voltage boost. The only thing you will get is a voltage drop by V=IR, even if you switch the current back and forth.

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So if I added a resistor to a circuit the voltage would increase and the current would decrease?

Ohm's law relates the voltage across to the current through a resistor

$$V_{R1} = I_{R1} \cdot R_1 $$

where the subscript R1 makes it explicit that the voltage and current variables in that equation are the voltage across the resistor and the current through the resistor. It occurs to me that you're trying to apply Ohm's law to a circuit as a whole.

We must use other circuit laws, e.g., KVL and KCL, to relate the resistor voltage and current to other circuit variables.

For example, consider the simple circuit formed by a battery and a resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

By KVL, the voltage across the resistor is fixed by the battery

$$V_{R1} = V_{BAT}$$

Also, since there is only one path for current, the series current is fixed by the resistor value via Ohm's law

$$I_S = I_{R1} = \frac{V_{R1}}{R_1} = \frac{V_{BAT}}{R_1}$$

Thus, in this circuit, changing the resistance \$R_1\$ will not change the voltage across the resistor since, we assume, the battery voltage is constant. However, the current through will change and is inversely proportional to the change in resistance.

Now, if we add another resistor in series

schematic

simulate this circuit

there are two resistor voltages to consider, \$V_{R1}\$ and \$V_{R2}\$ and, by KVL

$$V_{BAT} = V_{R1} + V_{R2}$$

which means

$$V_{R1} = V_{BAT} - V_{R2} < V_{BAT}$$

Thus, you see that adding the second resistor actually decreases the voltage across the first resistor as well as decreases the series current

$$I_S = I_{R1} = \frac{V_{R1}}{R_1} = \frac{V_{BAT} - V_{R2}}{R_1} < \frac{V_{BAT}}{R_1}$$

In summary, Ohm's law applies to the voltage across and current through the resistor and should not be applied carelessly to other circuit variables.

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Simplistically, resistors do not increase voltage, they reduce it. You only get "more" voltage if you have a resistor setup that uses up less of what you originally started with.

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  • \$\begingroup\$ Even simplistically, resistors don't inherently reduce voltage. If you place a simple 10-ohm resistor in series with a 10V source capable of delivering at least 1A, that source will continue to deliver 10V. \$\endgroup\$ – TDHofstetter Aug 26 '14 at 12:50
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It is not possible to increase the voltage using this law. Just understand the law properly. It states that Voltage is proportional to current. Resistance is the proportionality constant. SO if you have a certain amount of Voltage it can be reduced by using a resistor. But with the help of a resistor it is not possible to increase the voltage than the originally started voltage.

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  • \$\begingroup\$ V propotional to I \$\endgroup\$ – BASIL VARGHESE Aug 26 '14 at 10:47
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When looking at changes in values, Ohm's law tells you the relation between two of the circuit values when you keep the third one constant. V=IR tells you that for the same value of V, if I is higher in one circuit than another, then R is lower. Similarly, I = V/R tells you that for the same value of I, if V is higher in one circuit than another, then R is also higher. R=V/I is left as an exercise for the reader.

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