Question

There is a current of 192 pA when a certain potential is applied across a certain resistor. When that same potential is applied across a resistor made of the identical material but 28 times longer, the current is 0.054 pA. 1) Compare the effective diameters of the two resistors (i.e. find the ratio of the diameter of the second resistor to the diameter of the first resistor.). (Express your answer to two significant figures.)

Answer 1

Answer:

$\frac{d_2}{d_1}=0.089$

Explanation:

The equation that relates voltage, current and resistance is V=RI, and we can calculate this resistance in terms of the resistivity of a material $\rho$, its length L and cross-section (area) A with:

$R=\frac{\rho L}{A}$

Putting both equations together we have:

$\frac{V}{I}=\frac{\rho L}{A}$

Which will be useful to write as:

$\frac{V}{\rho}=\frac{I L}{A}$

since each term on the left will be the same for both situations, thus making the term on the right the same for both situations (which we will call 1 and 2), so we have:

$\frac{I_1 L_1}{A_1}=\frac{I_2 L_2}{A_2}$

Which we want to write as (since the ratio of the diameter is asked):

$\frac{A_2}{A_1}=\frac{I_2 L_2}{I_1 L_1}$

Since the areas appear to be that of a circle, we can write them as $A=\pi r^2=\frac{\pi d^2}{4}$ where r is the radius of that circle, and d the diameter. Substituting this in our previous equation we then have:

$\frac{A_2}{A_1}=\frac{\frac{\pi d_2^2}{4}}{\frac{\pi d_1^2}{4}}=\frac{d_2^2}{d_1^2}=(\frac{d_2}{d_1})^2=\frac{I_2 L_2}{I_1 L_1}$

Which means:

$\frac{d_2}{d_1}=\sqrt{\frac{I_2 L_2}{I_1 L_1}}$

We know that $I_1=192pA$, $I_2=0.054pA$ and $L_2=28L_1$, so substituting our values we have:

$\frac{d_2}{d_1}=\sqrt{\frac{(0.054pA)(28L_1)}{(192pA)(L_1)}}=0.089$

Answer 2

The ratio effective diameters of the resistors made of the same materials is 0.088.

Resistance of the circuit

V= IR

$R = \frac{\rho L }{A} \\\\V = \frac{I\rho L}{A} \\\\\frac{L_1I_1}{A_1} = \frac{L_2I_2}{A_2} \\\\\frac{A_2}{A_1} = \frac{L_2I_2}{L_1I_1}\\\\\frac{d_2^2}{d_1^2} = \frac{L_2I_2}{L_1I_1}\\\\\frac{d_2}{d_1}= \sqrt{ \frac{L_2I_2}{L_1I_1}}$

Comparing the effective diameters of the resistors

The ratio effective diameters of the resistors made of the same materials is calculated as follows;

$\frac{d_2}{d_1} = \sqrt{\frac{28l_1 \times 0.054p}{l_1 \times 192p} } \\\\\frac{d_2}{d_1} = 0.088$

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