A certain ohmic resistor has a resistance of 40 Ω. A second resistor is made of the same material but is three times as long and has half of the cross-sectional. What is the resistance of the second resistor? What is the effective resistance of the two resistors in series?

The resistance for the second resistor is R2 = 240 Ohms and the equivalent resistance is Requivalent = 280 Ohms.

Explanation:

The resistance of a ohmic resistor is influeced by the type of it's material and by the it's construction. The longer the wire the greater the resistance and the greater the cross-sectional the lower the resistance. This can be expressed by the following equation:

R = (p*L)/A

Where p is a constant for the material of the resistor, L is the length of the wire and A is the area of the cross-sectional. In our case we have a resistor R1 that has a resistance of 40 Ohms, while a second resistor R2 made with the same material but with double length and half cross sectional. If we say that R1 is:

R1 = (p*L)/A

Then R2 must be:

R2 = (p*3*L)/(A/2)

Because the only things that changed were the length and area of the cross-sectional. We can now relate both resistors to find the second resistance, using the equation for R2. So we have:

R2 = [3*(p*L)/A]*2 = 6*(p*L)/A = 6*R1

We know that R1 is 40 Ohms so R2 = 6*40 = 240 Ohms.

The equivalent resistance of a series connection is the sum of the individual resistances, so we have:

Requivalent = R1 + R2 = 40 + 240 = 280 Ohms.

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