Question

There is a current of 112 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 25 times longer, the current is 0.044 pA. Compare the effective diameters of the two resistors.

Answer 1

Answer:

First wire is 10 times larger in diameter than the second wire.

$\frac{ d_{1} }{ d_{2} }$ ≅ 10

Explanation:

Given :

Current in first wire $I_{1} = 112 \times 10^{-12}$ A

Current in second wire $I_{2} = 0.044 \times 10^{-12}$ A

Length of first wire = $l_{1}$

Length of second wire = $25 l_{1}$

Here potential across a resistor is same.

From ohm's law,

  $V =IR$

Where $R =$resistance of the wire,

Resistor of wire is given by,

  $R = \frac{\rho l}{A}$

Where $A =$ cross section area of the wire = $\pi r^{2}$

We write area in terms of diameter $d = 2r$

 $A = \frac{\pi d^{2} }{4}$

Now we have to compare current in both wires,

$I_{1} R_{1} = I_{2}R_{2}$

Put the value of resistor in terms of its diameter,

  $\frac{l_{1} d_{2}^{2} }{l_{2} d_{1}^{2} } = \frac{I_{2} }{I_{1} }$

$\frac{l_{1} d_{2}^{2} } { 25l_{1} d_{1}^{2} } = \frac{112 \times 10^{-12} }{0.044 \times 10^{-12} }$

     $\frac{ d_{2}^{2} }{ d_{1}^{2} } = 0.0098$

     $\frac{ d_{2} }{ d_{1}} } = \sqrt{0.0098} = 0.099$

     $\frac{ d_{1} }{ d_{2} } = \frac{1}{0.099} = 10.10$

     $\frac{ d_{1} }{ d_{2} }$ ≅ 10

So we can say that first wire is 10 times larger in diameter than the second wire.