30.10 LR Circuits

A circuit consisting of an inductor, a resistor, and (perhaps) a battery is called an LR circuit. FIGURE 30.45a is an example of an LR circuit. We’ll assume that the switch has been in position 1 for such a long time that the current is steady and unchanging. There’s no potential difference across the inductor, because $d I / d t = 0$ , so it simply acts like a piece of wire. The current flowing around the circuit is determined entirely by the battery and the resistor: $I_{0} = Δ V_{bat} / R$ .

The figure labeled (“a”) illustrates an L R circuit attached to a switch and a battery.

The figure labeled (b) illustrates a discharging L R circuit.

Full Alternative Text

What happens if, at $t = 0$ , the switch is suddenly moved to position 2? With the battery no longer in the circuit, you might expect the current to stop immediately. But the inductor won’t let that happen. The current will continue for some period of time as the inductor’s magnetic field drops to zero. In essence, the energy stored in the inductor allows it to act like a battery for a short period of time. Our goal is to determine how the current decays after the switch is moved.

NOTE It’s important not to open switches in inductor circuits because they’ll spark, as Figure 30.41 showed. The unusual switch in Figure 30.45 is designed to make the new contact just before breaking the old one.

FIGURE 30.45b shows the circuit after the switch is changed. Our starting point, once again, is Kirchhoff’s voltage law. The potential differences around a closed loop must sum to zero. For this circuit, Kirchhoff’s law is

Δ V_{res} + Δ V_{L} = 0

(30.53)

The potential differences in the direction of the current are $Δ V_{res} = - I R$ for the resistor and $Δ V_{L} = - L (d I / d t)$ for the inductor. Substituting these into Equation 30.53 gives

- R I - L \frac{d I}{d t} = 0

(30.54)

We’re going to need to integrate to find the current I as a function of time. Before doing so, we rearrange Equation 30.54 to get all the current terms on one side of the equation and all the time terms on the other:

\frac{d I}{I} = - \frac{R}{L} d t = - \frac{d t}{(L / R)}

(30.55)

We know that the current at $t = 0$ , when the switch was moved, was $I_{0}$ . We want to integrate from these starting conditions to current I at the unspecified time t. That is,

\int_{I_{0}}^{I} \frac{d I}{I} = - \frac{1}{(L / R)} \int_{0}^{t} d t

(30.56)

Both are common integrals, giving

ln I |_{I_{0}}^{I} = ln I - ln I_{0} = ln (\frac{I}{I_{0}}) = - \frac{t}{(L / R)}

(30.57)

We can solve for the current I by taking the exponential of both sides, then multiplying by $I_{0}$ . Doing so gives I, the current as a function of time:

I = I_{0} e^{- t / (L / R)}

(30.58)

Notice that $I = I_{0}$ at $t = 0$ , as expected.

The argument of the exponential function must be dimensionless, so $L / R$ must have dimensions of time. If we define the time constant $τ$ of the LR circuit to be

τ = \frac{L}{R}

(30.59)

then we can write Equation 30.58 as

I = I_{0} e^{- t / τ}

(30.60)

The time constant is the time at which the current has decreased to $e^{- 1}$ (about 37%) of its initial value. We can see this by computing the current at the time $t = τ :$

I (at t = τ) = I_{0} e^{- τ / τ} = e^{- 1} I_{0} = 0.37 I_{0}

(30.61)

Thus the time constant for an LR circuit functions in exactly the same way as the time constant for the RC circuit we analyzed in Chapter 29. At time $t = 2 τ$ , the current has decreased to $e^{- 2} I_{0}$ , or about 13% of its initial value.

The current is graphed in FIGURE 30.46. You can see that the current decays exponentially. The shape of the graph is always the same, regardless of the specific value of the time constant $τ$ .

The figure illustrates a current versus time graph of the current decay in an L R circuit.

FIGURE 30.46 Full Alternative Text

EXAMPLE 30.16 Exponential decay in an LR circuit

The switch in FIGURE 30.47 has been in position 1 for a long time. It is changed to position 2 at $t = 0 s$ .

The figure illustrates an L R circuit attached to a switch and a battery.

FIGURE 30.47 Full Alternative Text

What is the current in the circuit at $t = 5.0 μ s ?$
At what time has the current decayed to 1% of its initial value?

MODEL This is an LR circuit. We’ll assume ideal wires and an ideal inductor.

VISUALIZE The two resistors will be in series after the switch is thrown.

SOLVE Before the switch is thrown, while $Δ V_{L} = 0$ , the current is $I_{0} = (10 V)/ (100 Ω) = 0.10 A = 100 mA$ . This will be the initial current after the switch is thrown because the current through an inductor can’t change instantaneously. The circuit resistance after the switch is thrown is $R = 200 Ω$ , so the time constant is

τ = \frac{L}{R} = \frac{2.0 \times 10^{- 3} H}{200 Ω} = 1.0 \times 10^{- 5} s = 10 μ s

The current at $t = 5.0 μ s$ is

$I = I_{0} e^{- t / τ} = (100 mA) e^{- (5.0 μ s)/ (10 μ s)} = 61 mA$
To find the time at which a particular current is reached we need to go back to Equation 30.57 and solve for t:

$t = - \frac{L}{R} ln (\frac{I}{I_{0}}) = - τ ln (\frac{I}{I_{0}})$

The time at which the current has decayed to 1 mA $(1% of I_{0})$ is

$t = - (10 μ s) ln (\frac{1 mA}{100 mA}) = 46 μ s$

REVIEW For all practical purposes, the current has decayed away in $\approx 50 μ s$ . The inductance in this circuit is not large, so a short decay time is not surprising.

CHAPTER 30 CHALLENGE EXAMPLE Induction heating

Induction heating uses induced currents to heat metal objects to high temperatures for applications such as surface hardening, brazing, or even melting. To illustrate the idea, consider a copper wire formed into a $4.0 cm \times 4.0 cm$ square loop and placed in a magnetic field—perpendicular to the plane of the loop—that oscillates with 0.010 T amplitude at a frequency of 1000 Hz. What is the wire’s initial temperature rise, in °C/min?

MODEL The changing magnetic flux through the loop will induce a current that, because of the wire’s resistance, will heat the wire. Eventually, when the wire gets hot, heat loss through radiation and/or convection will limit the temperature rise, but initially we can consider the temperature change due only to the heating by the current. Assume that the wire’s diameter is much less than the 4.0 cm width of the loop.

VISUALIZE FIGURE 30.48 shows the copper loop in the magnetic field. The wire’s cross-section area A is unknown, but our assumption of a thin wire means that the loop has a well-defined area $L^{2}$ . Values of copper’s resistivity, density, and specific heat were taken from tables inside the back cover of the book. We’ve used subscripts to distinguish between mass density $ρ_{mass}$ and resistivity $ρ_{elec}$ , a potentially confusing duplication of symbols.

The figure illustrates induction by heating a copper wire forming a square loop.

FIGURE 30.48 Full Alternative Text

SOLVE Power dissipation by a current, $P = I^{2} R$ , heats the wire. As long as heat losses are negligible, we can use the heating rate and the wire’s specific heat c to calculate the rate of temperature change. Our first task is to find the induced current. According to Faraday’s law,

I = \frac{ε}{R} = - \frac{1}{R} \frac{d Φ_{m}}{d t} = - \frac{L^{2}}{R} \frac{d B}{d t}

where $R$ is the loop’s resistance and $Φ_{m} = L^{2} B$ is the magnetic flux through a loop of area $L^{2}$ . The oscillating magnetic field can be written $B = B_{0} cos ω t$ , with $B_{0} = 0.010 T$ and $ω = 2 π \times 1000 Hz = 6280 rad / s$ . Thus

\frac{d B}{d t} = - ω B_{0} sin ω t

from which we find that the induced current oscillates as

I = \frac{ω B_{0} L^{2}}{R} sin ω t

As the current oscillates, the power dissipation in the wire is

P = I^{2} R = \frac{ω^{2} B_{0}^{2} L^{4}}{R} {sin}^{2} ω t

The power dissipation also oscillates, but very rapidly in comparison to a temperature rise that we expect to occur over seconds or minutes. Consequently, we are justified in replacing the oscillating $P$ with its average value $P_{avg}$ . Recall that the time average of the function ${sin}^{2} ω t$ is $\frac{1}{2}$ , a result that can be proven by integration or justified by noticing that a graph of ${sin}^{2} ω t$ oscillates symmetrically between 0 and 1. Thus the average power dissipation in the wire is

P_{avg} = \frac{ω^{2} B_{0}^{2} L^{4}}{2 R}

Recall that power is the rate of energy transfer. In this case, the power dissipated in the wire is the wire’s heating rate: $d Q / d t = P_{avg}$ , where here $Q$ is heat, not charge. Using $Q = m c Δ T$ , from thermodynamics, we can write

\frac{d Q}{d t} = m c \frac{d T}{d t} = P_{avg} = \frac{ω^{2} B_{0}^{2} L^{4}}{2 R}

To complete the calculation, we need the mass and resistance of the wire. The wire’s total length is $4 L$ , and its cross-section area is $A$ . Thus

\begin{array}{l} m & = & ρ_{mass} V = 4 ρ_{mass} L A \\ R & = & \frac{ρ_{elec} (4 L)}{A} = \frac{4 ρ_{elec} L}{A} \end{array}

Substituting these into the heating equation, we have

4 ρ_{mass} L A c \frac{d T}{d t} = \frac{ω^{2} B_{0}^{2} L^{3} A}{8 ρ_{elec}}

Interestingly, the wire’s cross-section area cancels. The wire’s temperature initially increases at the rate

\frac{d T}{d t} = \frac{ω^{2} B_{0}^{2} L^{2}}{32 ρ_{elec} ρ_{mass} c}

All the terms on the right-hand side are known. Evaluating, we find

\frac{d T}{d t} = 3.3 K / s = 200 ° C / min

REVIEW This is a rapid but realistic temperature rise for a small object, although the rate of increase will slow as the object begins losing heat to the environment through radiation and/or convection. Induction heating can increase an object’s temperature by several hundred degrees in a few minutes.

30.10 LR Circuits

FIGURE 30.45 An LR circuit.

FIGURE 30.46 The current decay in an LR circuit.

EXAMPLE 30.16 Exponential decay in an LR circuit

FIGURE 30.47 The LR circuit of Example 30.16.

CHAPTER 30 CHALLENGE EXAMPLE Induction heating

FIGURE 30.48 A copper wire being heated by induction.