When you connect an unknown resistor across the terminals of a 1.50 V AAA batter

Question

When you connect an unknown resistor across the terminals of a 1.50 V AAA battery having negligible internal resistance, you measure a current of 18.3 mA flowing through it.

Part A

What is the resistance of this resistor?

R = ___

Part B

If you now place the resistor across the terminals of a 12.3 V car battery having no internal resistance, how much current will flow?

I1 = ___ mA

Part C

You now put the resistor across the terminals of an unknown battery of negligible internal resistance and measure a current of 0.451 A flowing through it. What is the potential difference across the terminals of the battery?

V2 = ___V

Part D

If you triple the length of a cable and at the same time triple its diameter, what will be its resistance if its original resistance was R?

Express your answer using three significant figures.

R1/R = _______ .....( I put 3/4, and 0.75 and it says it wrong)

Explanation / Answer

A) Apply Ohm's law

V = I*R

==> R = V/I

= 1.5/(18.3*10^-3)

= 81.97 ohms

B) I = V/R

= 12.3/8197

= 150 mA

C) V = I*R

= 0.451*81.97

= 36.97 Volts

D) we know, R = rho*L/A

= rho*L/(pi*d^2/4)

so, when triple the length of a cable and at the same time triple its diameter

R1 = rho*3*L/(pi*(3*d)^2/4)

= (1/3)*rho*L/(pi*(d/2)^2)

= R/3

R1/R = 1/3 = 0.333

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