Question

Define the inductive time constant of an LR circuit. An inductance coil L, a resistor R and a battery are connected in series. The time constant of the circuit is 3.5 x 10^-3 s.
When a resistance of 6 ohm is added in series, a new time constant of 1.5 x 10^-3 s is obtained. Calculate the inductance of the coil and the resistance of the resistor.

Answer 1

$$3.5 \cdot 10^{-3}=\tau_1=\frac{L}{R_1}$$

$$1.5 \cdot 10^{-3}=\tau_2=\frac{L}{R_1+R_2}, \quad R_2=6\, ohm$$

$$\tau_2 (R_1+R_2)=\tau_1 R_1 \Rightarrow R_1=\frac{\tau_2 R_2}{\tau_1-\tau_2}=4.5\,ohm$$

$$L=\tau_1 R_1=15.75 \cdot 10^{-3}$$