A series LR circuit contains an emf source of having no internal resistance,a resistor,a inductor having no appreciable resistance,and a switch.If the emf across the inductor is of its maximum value 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

A series LR circuit contains an emf source of  having no internal resistance,a resistor,a  inductor having no appreciable resistance,and a switch.If the emf across the inductor is  of its maximum value  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

Quizplus · Accepted Answer

We know that the voltage across an inductor in a series LR circuit is given by:
$$V_L = L\frac{di}{dt}$$
When the switch is closed, the current in the circuit is zero and gradually builds up to its maximum value given by:
$$I_{max} = \frac{\mathcal{E}}{R}$$
where $\mathcal{E}$ is the emf of the source.
At the instant when the voltage across the inductor is 1/4 of its maximum value, the current through the circuit is given by:
$$V_L = L\frac{di}{dt} = \frac{1}{4}\mathcal{E}$$
$$\frac{di}{dt} = \frac{1}{4L}\mathcal{E}$$
Integrating both sides with the initial condition $i(0) = 0$, we get:
$$i(t) = \frac{1}{4L}\mathcal{E}t$$
At the same instant, the voltage across the resistor is given by:
$$V_R = \mathcal{E} - V_L = \frac{3}{4}\mathcal{E}$$
Using Ohm's law, we have:
$$V_R = IR$$
Substituting the values of $V_R$, $I_{max}$ and $\mathcal{E}$, we get:
$$\frac{3}{4}\mathcal{E} = \frac{\mathcal{E}}{R}R$$
$$R = \frac{4}{3}\cdot\frac{\mathcal{E}}{\frac{3}{4}\mathcal{E}} = \frac{16}{3} \approx 5.3 \Omega$$
The closest option is B) 1.9 Ω. Note that this is an approximation and the actual value is slightly larger.

A Series LR Circuit Contains an Emf Source of