Question

A carbon resistor is 5 mm long and hasa constant cross section of 0.5mm2. The conductivity of carbon at room temperature is = 3104 per ohm·m. In a circuit itspotential at one end of the resistor is 13 volts relative to ground, and at the other endthe potential is 17 volts. Calculatethe resistance R and the current I.R = .333correct check mark0.333I =12.01amperesexplane how to work this problem with the actualwork.

Answer 1

Step-by-step explanation:

The given data is as follows.

l = 5 mm =
`5 * 10^(-3)` m, A = 0.5
`mm^(2)` =
`0.5 * 10^(-6) m^(2)`

conductivity (
`\sigma`) =
`3 * 10^(4)` per ohm-m

`V_(1)` = 13 volts,
`V_(2)` = 17 volts

Now, the relation between conductivity and resistivity is as follows.

`\sigma = (1)/(\rho)`

where,
`\rho` = resistivity

Therefore, calculate resistivity as follows.

`\sigma = (1)/(\rho)`

`\rho = (1)/(3 * 10^(4))`

=
`3.33 * 10^(-5)` ohm-m

Now, we will calculate the resistance using the following formula.

R =
`\rho (l)/(A)`

=
`3.33 * 10^(-5) ohm-m * (5 * 10^(-3) m)/(0.5 * 10^(-6) m^(2))`

= 0.333 ohm

Hence, value of resistance is 0.333 ohm.

As we know that relation between current and voltage is as follows.

V = IR

Therefore, calculate the value of current as follows.

V = IR

(17 - 13) =
`I * 0.333 ohm`

I = 12.01 ampere

Thus, value of the current is 12.01 A.