Question

What length of copper wire (resistivity 1.68x10-8\Omegam) of diameter 0.15 mm is needed for a total resistance of 15\Omega? if 6.00 Ohm resistor is connected to a 12.0 V battery with this wire, what would the voltage drop accross the resistor be? What would the voltage drop accross the wire be?

Answer 1

The length of copper wire required for a resistance of 15 Ω can be calculated using the area of the wire's cross-section and the formula R=(resistivity x length)/area. With a 6 Ohm resistor and a 12 V battery, the voltage drops can be calculated using Ohm's law to find the current first and then apply V=IR for the resistor and wire separately.

Step-by-step explanation:

To find the length of copper wire needed for a total resistance of 15 Ω, we first calculate the area of the cross-section of the wire using the diameter given. The diameter of the wire is 0.15 mm, which is 0.15 x 10-3 meters. The area A is then (π/4) x (diameter)2 = (π/4) x (0.15 x 10-3)2 square meters. Using the formula for resistance R = (resistivity x length) / area, we can rearrange to find the length, L = (R x A) / resistivity. Substituting the given values, the length L = (15 Ω x (π/4) x (0.15 x 10-3)2) / (1.68 x 10-8 Ωm).

When a 6.00 Ω resistor is connected to a 12.0 V battery along with this wire, the total resistance of the circuit is the sum of the resistance of the wire and the resistor. The voltage drop across the resistor can be calculated using Ohm's law, V = IR, where I is the current. The current I through the circuit can be found using the total resistance and the voltage of the battery, I = V / (Rtotal). Once we have the current, the voltage drop across the resistor VR is simply I x Rresistor, and the voltage drop across the wire Vwire is I x Rwire.

Answer 2

Answer: (a) The length of the wire with a diameter of 0.15m and 15-ohm resistance is 15.77 m

(b) There is no voltage drop in the resistor but there is a voltage drop of 2 V in the wire

Step-by-step explanation:

The expression for resistance is R = xL/A, where R is resistance, x is Resistivity, L is the length of the wire and A is the area

Rearranging the equation we get, L = RA/x

The area of a cross-section is A =
`\pi`(D/2)^2

D in our case is equal to 0.15 x 10^-3 m

x in our case is equal to 1.68 x 10^-8 omega m

Putting all these values in the equation we get L to be equal to 15.77 m

For the second part using our equation V = IR, we can figure out there will be no voltage drop across the resistor but there will be a voltage drop across the wire by 2 V

That is due to the new resistor being added in a series connection and using that information we can find out the current in the wire, as the current stays the same across resistors in a series connection we can find out the voltage in each resistor using our equation V = IR

Subtracting it from the initial voltage through the wire will give us the change i.e. of 2 V