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

Explanation:

The given data is as follows.

       l = 5 mm = $5 \times 10^{-3}$ m,   A = 0.5 $mm^{2}$ = $0.5 \times 10^{-6} m^{2}$

 conductivity ($\sigma$) = $3 \times 10^{4}$ per ohm-m

      $V_{1}$ = 13 volts,       $V_{2}$ = 17 volts

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

              $\sigma = \frac{1}{\rho}$

where,     $\rho$ = resistivity

Therefore, calculate resistivity as follows.

                   $\sigma = \frac{1}{\rho}$

                  $\rho = \frac{1}{3 \times 10^{4}}$  

                            = $3.33 \times 10^{-5}$ ohm-m          

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

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

                      = $3.33 \times 10^{-5} ohm-m \times \frac{5 \times 10^{-3} m}{0.5 \times 10^{-6} m^{2}}$

                      = 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) = $I \times 0.333 ohm$

                    I = 12.01 ampere

Thus, value of the current is 12.01 A.