Chapter 14: InductanceContent check — University Physics 2, Chapter 14: Inductance

Describe how the currents through $R_1\text{and}{R}_2$ shown below vary with time after switch S is closed.

For the circuit shown below, $\epsilon =20\text{V}$, $L=4.0\text{mH,}$ and $R=5.0\text{Ω}$. After steady state is reached with ${\text{S}}_1$ closed and ${\text{S}}_2$ open, ${\text{S}}_2$ is closed and immediately thereafter $(\text{at}t=0)$ ${\text{S}}_1$ is opened. Determine (a) the current through L at $t=0$, (b) the current through L at $t=4.0\times {10}^{\mathrm{-4}}\text{s}$, and (c) the voltages across L and $R_2$ at $t=4.0\times {10}^{\mathrm{-4}}\text{s}$. $R_1=R_2=R$.

How long after switch ${\text{S}}_1$ is thrown does it take the current in the circuit shown to reach half its maximum value? Express your answer in terms of the time constant of the circuit.

The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch ${\text{S}}_1$ is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if $L=250\text{mH}$.

For the circuit shown below, $\epsilon =50\text{V}$, $R_1=10\text{Ω,}$, $R_2=$ $R_3=$ $\text{19.4 Ω,}$, and $L=2.0\text{mH}$. Find the values of $I_1\text{and}{I}_2$ (a) immediately after switch S is closed, (b) a long time after S is closed, (c) immediately after S is reopened, and (d) a long time after S is reopened.

Show that for the circuit shown below, the initial energy stored in the inductor, $L{I}^2(0)\text{/}2$, is equal to the total energy eventually dissipated in the resistor, ${\displaystyle {\int }_0^{\infty }{I}^2(t)}Rdt$.

A small, rectangular single loop of wire with dimensions l, and a is placed, as shown below, in the plane of a much larger, rectangular single loop of wire. The two short sides of the larger loop are so far from the smaller loop that their magnetic fields over the smaller fields over the smaller loop can be ignored. What is the mutual inductance of the two loops?

The switch S of the circuit shown below is closed at $t=0$. Determine (a) the initial current through the battery and (b) the steady-state current through the battery.

In a damped oscillating circuit the energy is dissipated in the resistor. The Q-factor is a measure of the persistence of the oscillator against the dissipative loss. (a) Prove that for a lightly damped circuit the energy, U, in the circuit decreases according to the following equation.

$\frac{dU}{dt}=\mathrm{-2}\beta U,$ where $\beta =\frac{R}{2L}.$

(b) Using the definition of the Q-factor as energy divided by the loss over the next cycle, prove that Q-factor of a lightly damped oscillator as defined in this problem is $$

$Q\equiv \frac{U_{\text{begin}}}{\text{Δ}{U}_{\text{one cycle}}}=$ $\frac{1}{2\pi R}\sqrt{\frac{L}{C}}.$

(Hint: For (b), to obtain Q, divide E at the beginning of one cycle by the change $\text{Δ}E$ over the next cycle.)

A square loop of side 2 cm is placed 1 cm from a long wire carrying a current that varies with time at a constant rate of 3 A/s as shown below. (a) Use Ampère’s law and find the magnetic field. (b) Determine the magnetic flux through the loop. (c) If the loop has a resistance of $3\text{Ω}$, how much induced current flows in the loop?