Circuits with Resistors and Inductors
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AP Physics C: Electricity and Magnetism › Circuits with Resistors and Inductors
What is the physical significance of the time constant, $$\tau = L/R$$, for this circuit?
It is the time at which the energy stored in the inductor is half of its maximum value.
It is the time required for the current to reach approximately 63% of its final steady-state value.
It is the time at which the potential difference across the inductor is equal to the potential difference across the resistor.
It is the time required for the current to reach its maximum steady-state value.
Explanation
The time constant $$\tau = L/R$$ in a charging LR circuit is defined as the time it takes for the current to rise to $$1 - 1/e$$ of its final value, which is approximately 63%. The current approaches its maximum value asymptotically and never truly reaches it in finite time. The other conditions occur at different times, not necessarily at $$t=\tau$$.
What is the primary reason for this delay in the current reaching its maximum value?
The internal resistance of the battery limits the initial rate at which charge can be supplied to the circuit.
The capacitance of the connecting wires stores charge, which must be built up before current can flow steadily.
The resistor dissipates energy, which slows down the flow of charge carriers through the circuit.
The inductor generates a back emf that opposes the increase in current, a property known as inductance.
Explanation
The defining property of an inductor is its inductance, which is its tendency to oppose a change in current. When the switch is closed, the current tries to increase from zero. This change in current creates a changing magnetic flux in the inductor, which in turn induces a back emf that opposes the current's increase. This opposition is what causes the gradual, exponential rise in current rather than an instantaneous one.
Which of the following describes the current $$I(t)$$ in the resistor for $$t > 0$$?
The current immediately drops to zero as the energy source is removed.
The current remains constant at $$I_0$$ because the inductor resists any change.
The current decays exponentially toward zero with a time constant of $$L/R$$.
The current decays linearly to zero over a time interval of $$L/R$$.
Explanation
When the battery is removed, the inductor's stored magnetic energy becomes the source for the circuit. The inductor drives a current that initially has the value $$I_0$$. This energy is dissipated in the resistor, causing the current to decrease. The decay follows an exponential function $$I(t) = I_0 e^{-t/\tau}$$, where the time constant is $$\tau = L/R$$.
At what rate is energy being stored in the magnetic field of the inductor at time $$t=0$$?
At an infinite rate, because the change in energy is instantaneous at the start.
At a rate of $$E^2/R$$, which is the maximum power delivered by the battery.
At a rate of $$E^2/L$$, derived from the initial rate of current change.
At a rate of zero, because the current is zero.
Explanation
The energy stored in an inductor is $$U_L = \frac{1}{2}LI^2$$. The rate at which energy is stored is the power, $$P_L = \frac{dU_L}{dt} = \frac{d}{dt}(\frac{1}{2}LI^2) = LI \frac{dI}{dt}$$. At $$t=0$$, the current $$I$$ is zero. Therefore, the rate of energy storage $$P_L$$ is also zero, even though $$dI/dt$$ is at its maximum value of $$E/L$$.
After a very long time ($$t \to \infty$$), what is the potential difference across the inductor?
It depends on the value of $$L/R$$, which determines the final state of the inductor.
Zero, because the current has reached a constant maximum value, so its rate of change is zero.
$$E$$, because the inductor stores the full energy provided by the battery's emf.
$$E/2$$, because the inductor's impedance matches the resistor's resistance at steady state.
Explanation
After a long time, the circuit reaches a steady state where the current is constant at its maximum value, $$I = E/R$$. The potential difference across the inductor is given by $$V_L = L \frac{dI}{dt}$$. Since the current is constant, $$\frac{dI}{dt} = 0$$, and therefore the potential difference across the inductor is zero. The inductor behaves like a simple connecting wire in a DC circuit at steady state.
What is the total energy dissipated by the resistor as the current decays from $$I_0$$ to zero?
Zero, because the energy returns to the inductor.
$$\frac{1}{2}L I_0^2$$
$$L R I_0$$
$$\frac{1}{2}R I_0^2$$
Explanation
Initially, at steady state, the inductor stores an amount of energy equal to $$U_L = \frac{1}{2}L I_0^2$$. When the battery is removed, this stored energy is the only energy in the circuit. As the current decays, this energy is entirely dissipated as heat in the resistor. By conservation of energy, the total energy dissipated by the resistor must equal the initial energy stored in the inductor.
What is the time constant of this circuit after the switch is closed?
$$\frac{L(R_1+R_2)}{R_1R_2}$$
$$L \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$
$$\frac{L R_1 R_2}{R_1+R_2}$$
$$\frac{L}{R_1 + R_2}$$
Explanation
The time constant of an LR circuit is given by $$\tau = L/R_{eq}$$, where $$R_{eq}$$ is the equivalent resistance seen by the inductor. In this circuit, the two resistors $$R_1$$ and $$R_2$$ are in parallel. Their equivalent resistance is $$R_{eq} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = \frac{R_1R_2}{R_1+R_2}$$. Therefore, the time constant is $$\tau = \frac{L}{R_{eq}} = \frac{L}{\frac{R_1R_2}{R_1+R_2}} = \frac{L(R_1+R_2)}{R_1R_2}$$.
If the resistance is doubled to $$2R_0$$ and the inductance is halved to $$L_0/2$$, what is the new time constant $$\tau_{new}$$?
$$\tau_0 / 4$$
$$4\tau_0$$
$$\tau_0$$
$$\tau_0 / 2$$
Explanation
The time constant of an LR circuit is given by the formula $$\tau = L/R$$. The initial time constant is $$\tau_0 = L_0/R_0$$. The new time constant is $$\tau_{new} = (L_0/2) / (2R_0) = L_0 / (4R_0) = (1/4)(L_0/R_0) = \tau_0/4$$.
When determining the time constant for the decay of current through the inductor after the battery has been disconnected for a long time (by shorting the terminals where the battery was), what is the effective resistance used in the calculation of $$\tau$$?
$$\frac{R_1 R_2}{R_1 + R_2}$$, because the resistors are in parallel with respect to the inductor during decay.
$$R_1 + R_2$$, because the current from the inductor flows through both resistors.
$$R_1$$, because it is in the same branch as the inductor.
$$R_2$$, because it provides the alternate path for the current.
Explanation
When the battery is removed and its terminals are shorted, the inductor and resistors form a single loop for the current to decay. The current from the inductor must flow through both $$R_1$$ and $$R_2$$ in series to complete the loop. Therefore, the equivalent resistance that governs the decay is the series combination, $$R_{eq} = R_1 + R_2$$. The time constant would be $$\tau = L/(R_1 + R_2)$$.
A series $R=4,\Omega$, $L=2,\text{H}$ is connected to $8,\text{V}$ at $t=0$; Refer to the scenario above. How does the current change over time after the switch is closed?
$i(t)=2\left(1-e^{-2t}\right),\text{A}$
$i(t)=2\left(1-e^{-t/2}\right),\text{A}$
$i(t)=2\left(1-e^{+2t}\right),\text{A}$
$i(t)=2e^{-2t},\text{A}$
Explanation
This question tests AP Physics C skills in understanding LR circuits and electromagnetic induction principles. When a switch closes in an RL circuit, the current grows exponentially according to i(t) = (V/R)(1 - e^(-t/τ)), where τ = L/R is the time constant. In this scenario, V = 8 V, R = 4 Ω, and L = 2 H, giving τ = 2/4 = 0.5 s and steady-state current I₀ = V/R = 8/4 = 2 A. Choice A is correct because i(t) = 2(1 - e^(-t/0.5)) = 2(1 - e^(-2t)) A, showing exponential growth toward 2 A. Choice B incorrectly represents decay rather than growth, while C has the wrong exponent, and D has a positive exponent which would cause unbounded growth. To help students: Remember that closing a switch causes current growth (1 - e^(-t/τ)) while opening causes decay (e^(-t/τ)). Always verify that the exponent is negative and equals -t/τ = -tR/L.