Suppose we have a 5 volt, or 5 J / C potential difference across equal lengths of a thick wire and thin wire.

• We can produce this potential difference by cranking the generator at the appropriate constant rate.
• There are more charge carriers in the thick wire, so it turns out that there will be a greater current in the thick wire.
• More current implies that there is more power dissipated in the thick wire, requiring more power from the generator.
• Since the rate of cranking is constant, to produce this power we must therefore have a greater force at the generator.

If we have a shorter wire and a longer wire, both of the same diameter and both with the same 5 J / C potential difference across them, the longer wire will have more charge carriers, which would tend to result in a greater current, but a lower potential gradient, which tends to produce less current.

• We therefore have two contrasting influences, which might cancel out, though one influence might be greater than the other.

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Suppose that we have a current of .2 amps, or .2 C / s, passing through a circuit containing two bulbs.

• The bulbs are connected in series, meaning that what flows through one bulb must flow through the other.
• The thin wire resists the flow of current more than the thick wire, and hence requires more voltage to produce the same current.
• We define the resistance of a bulb or another circuit element to be the ratio of voltage to current, R = V / I, through the device.
• A high resistance thus implies that we require more voltage for given current, or alternatively that we get less current for a given voltage.

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When we have to resisting devices, for example two bulbs, connected in series, as we have seen the same current must flow through both.

• Since resistance has been defined as R = V / I, we see that when a current I flows across a resistance R, there must be a voltage V = I R across that resistance.
• If I is the current, then the voltage V1 across the resistance R1 is I * R1; the voltage across R2 is V2 = I * R2.
• We see that the total voltage across the circuit is therefore V1 + V2 = I * R1 + I * R2 = I * (R1 + R2) = I * R, where R = R1 + R2 is the equivalent resistance for the two resistors.