A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor?
A) 14 Ω
B) 1.9 Ω
C) 1.5 Ω
D) 5.0 Ω

Question

Quizplus · Accepted Answer

The time constant of the circuit is given by $	au = L/R$, where $L$ is the inductance and $R$ is the resistance. At $t = 0$, the voltage across the inductor is zero, and it increases exponentially towards its maximum value, according to:

$$
V_L = \mathcal{E}(1 - e^{-t/	au})
$$

At $t = 4.0\,	ext{s}$, the voltage across the inductor is $0.8\,\mathcal{E}$. Therefore, we can write:

$$
0.8\,\mathcal{E} = \mathcal{E}(1 - e^{-4.0/	au})
$$

Simplifying this expression, we get:

$$
e^{-4.0/	au} = 0.2
$$

Taking the natural logarithm of both sides, we get:

$$
-\frac{4.0}{	au} = \ln 0.2 \quad \Rightarrow \quad 	au = \frac{4.0}{\ln 5}
$$

Now we can solve for the resistance $R$:

$$
R = \frac{L}{	au} = \frac{34}{4.0/ \ln 5} \approx 1.9\,\Omega
$$

Therefore, the answer is B).

A Series LR Circuit Contains an Emf Source of 14