14.4 Rl circuits  (Page 3/4)

Use Lenz’s law to explain why the initial current in the RL circuit of [link] (b) is zero.

When the current in the RL circuit of [link] (b) reaches its final value $\text{ε}\text{/}R,$ what is the voltage across the inductor? Across the resistor?

Does the time required for the current in an RL circuit to reach any fraction of its steady-state value depend on the emf of the battery?

An inductor is connected across the terminals of a battery. Does the current that eventually flows through the inductor depend on the internal resistance of the battery? Does the time required for the current to reach its final value depend on this resistance?

At what time is the voltage across the inductor of the RL circuit of [link] (b) a maximum?

In the simple RL circuit of [link] (b), can the emf induced across the inductor ever be greater than the emf of the battery used to produce the current?

If the emf of the battery of [link] (b) is reduced by a factor of 2, by how much does the steady-state energy stored in the magnetic field of the inductor change?

A steady current flows through a circuit with a large inductive time constant. When a switch in the circuit is opened, a large spark occurs across the terminals of the switch. Explain.

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

Discuss possible practical applications of RL circuits.

In [link] , $\epsilon =12\text{V}$ , $L=20\text{mH}$ , and $R=5.0\text{Ω}$ . Determine (a) the time constant of the circuit, (b) the initial current through the resistor, (c) the final current through the resistor, (d) the current through the resistor when $t=2{\tau }_L,$ and (e) the voltages across the inductor and the resistor when $t=2{\tau }_L.$

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 at $t=4.0\times {10}^{\mathrm{-4}}\text{s}$ .

The current in the RL circuit shown here increases to $40\text{\%}$ of its steady-state value in 2.0 s. What is the time constant of the circuit?

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.

Examine the circuit shown below in part (a). Determine dI/dt at the instant after the switch is thrown in the circuit of (a), thereby producing the circuit of (b). Show that if I were to continue to increase at this initial rate, it would reach its maximum $\epsilon \text{/}R$ in one time constant.

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 $L=250\text{mH}$ .

Consider the circuit shown below. Find $I_1,I_2,\text{and}{I}_3$ when (a) the switch S is first closed, (b) after the currents have reached steady-state values, and (c) at the instant the switch is reopened (after being closed for a long time).

For the circuit shown below, $\epsilon =50\text{V}$ , $R_1=10\text{Ω,}$ 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.

For the circuit shown below, find the current through the inductor $2.0\times {10}^{\mathrm{-5}}\text{s}$ after the switch 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, ${\displaystyle {\int }_0^{\infty }{I}^2(t)}Rdt$ .