Question

A 13.0-V battery is connected in series with a switch, resistor and coil. If the circuit's time constant is 1.20 ´ 10-4 s and the final steady current after the switch is closed becomes 2.50 A

Answer 1

Answer:

R = 5.20 Ω,    L = 6.24 10⁻⁴ H

Explanation:

The current in an RL circuit is

           I = $\frac{E}{R} \ (1- e^{- t/ \tau } )$

           τ = L / R

In the problem they indicate the value of the voltage, the current and the time constant, for which the resistance must be found

         

The stable current is when enough time has passed (t »τ) after closing the circuit, therefore the exponential term is very small and we can neglect it.

            I = E / R

            R = E / I

let's calculate

            R = 13.0 / 2.50

            R = 5.20 Ω

now with this value we can find the inductance of the coil

            τ = L / R

            L = τ  R

            L = 1.20 10⁻⁴   5.2

            L = 6.24 10⁻⁴ H