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Chapter 24: Problem 8

You put a \(1.5-\mathrm{V}\) battery across a piece of material, and a \(100-\mathrm{mA}\) current flows. With a 9 - \(V\) battery, the current increases to 400 mA. Is the material ohmic or not?

Short Answer

Expert verified
The material is not ohmic as the resistance changes when the voltage (or current) changes.

Step by step solution

01

Calculate Resistance for the First Case

First, calculate the resistance \( R1 \) when a \(1.5-V\) battery is used which results in a current of \(100-mA\) (you have to take this into account as \(0.100-A\)). Resistance \( R1 \) can be calculated by using Ohm's law \(V = I*R\). Solving for \(R\), we get \(R = V/I\). Substitute \(V = 1.5-V\) and \(I = 0.100-A\) into the equation. The resistance \(R1\) will then be \( R1 = 1.5/0.100 = 15 \Omega\).
02

Calculate Resistance for the Second Case

Next, calculate the resistance \( R2 \) when a \(9-V\) battery is used which results in a current of \(400-mA\) (take this as \(0.400-A\)). Again using the Ohm's law as in the previous step but changing the values for \(V\) and \(I\) to find resistance \(R2 = V/I = 9/0.400 = 22.5 \Omega\).
03

Compare the Resistances

Now compare the values of \(R1\) and \(R2\). If they are equal, the material is ohmic (as the resistance which is the ratio of voltage to current remains constant). If they are not equal, the material is not ohmic. In this case, \(R1\) is \(15 \Omega\) and \(R2\) is \( 22.5 \Omega\), so they are not equal.

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Most popular questions from this chapter

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