Question

During a summer internship as an electronics technician, you are asked to measure the self-inductance L of a solenoid. You connect the solenoid in series with a 10.0 Ω resistor, a battery that has negligible internal resistance, and a switch. Using an ideal voltmeter, you measure and digitally record the voltage v_L across the solenoid as a function of the time t that has elapsed since the switch is closed. Your measured values are shown in Figure below, where v_L is plotted versus t. In addition, you measure that v_L = 50.0 V just after the switch is closed and v_L = 20.0 V a long time after it is closed. (a) Apply the loop rule to the circuit and obtain an equation for v_L as a function of t. (b) What is the emf E of the battery? (c) According to your measurements, what is the voltage amplitude across the 10.0 Ω resistor as t → ∞? Use this result to calculate the current in the circuit as t → ∞. (d) What is the resistance R_L of the solenoid? (e) Use the theoretical equation from part (a), Fig below, and the values of E and R_L from parts (b) and (d) to calculate L.

Short answer

1. the variation of voltage across solenoid is represented by equation:

$V_L=i_0{R}_L=\frac{\epsilon R_L}{\left(R_0+R\right)}\left(1-e^{-\left(\frac{R_0+R}{L}\right)t}\right)$

b. the EMF of the battery is 50 volts.

c. the voltage drop across resistor is 30 volts and the current through resistor is 3 amperes.

d. the resistance of the solenoid is 6.7 ohms.

e. the self-inductance of the coil is 38.41 milli-henry.

Step-by-step solution

Basic Definition

Faraday’s law states that a current is induced in a conductor when it is exposed to a time varying magnetic flux. This induced current is driven by a force called electromotive or electromagnetic force. The magnitude of induced emf is given by

$\epsilon =-L\frac{di}{dt}$

Where L is the inductance of the conductor.

Lenz further explained the direction of this induced current. According to lenz, the direction of induced current will be such that the magnetic field created by the induced current opposes the changing magnetic field which caused its induction.

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. An inductor stores the energy in form of magnetic field.

A capacitor is a device consisting of two conductors in close proximity that are used to store electrical energy. A capacitor stores energy in form of electric field.

An LC-circuit can store energy while oscillating at its natural resonant frequency. For such circuits, the energy in the circuit shifts between electric and magnetic fields. The angular frequency of this oscillation is given by

$\omega =2\mathrm{\pi }f=\frac{1}{\sqrt{LC}}$

Voltage across solenoid

Apply kirchhoff’s rule on the bottom loop, we get

$\begin{array}{r}\epsilon -v_x-v_{cb}=0\\ \epsilon -i{R}_0-iR-L\frac{di}{dt}=0\\ L\frac{di}{dt}=\epsilon -i\left(R_0+R\right)\\ \frac{di}{-i+\epsilon /\left(R_0+R\right)}=\left(\frac{R_0+R}{L}\right)dt\end{array}$

Integrating both sides from t_i = 0 to t_f = t sec

$\begin{array}{r}\ln \left(\frac{-i_0+\epsilon /\left(R_0+R\right)}{\epsilon /\left(R_0+R\right)}\right)=-\left(\frac{R_0+R}{L}\right)t\\ \frac{-i_0+\epsilon /\left(R_0+R\right)}{\epsilon /\left(R_0+R\right)}=e^{-\left(\frac{R_0+R}{L}\right)t}\\ i_0=\frac{\epsilon }{\left(R_0+R\right)}\left(1-e^{-\left(\frac{R_0+R}{L}\right)t}\right)\end{array}$

Using this current variation, voltage drop across inductor is given by

$V_L=i_0{R}_L=\frac{\epsilon R_L}{\left(R_0+R\right)}\left(1-e^{-\left(\frac{R_0+R}{L}\right)t}\right)$

Therefore, the variation of voltage across solenoid is represented using above equation.

EMF of battery

As the voltage drop across inductor right after closing the switch is 50 volts and there’s no drop of voltage anywhere else in the circuit.

Therefore, the EMF of the battery is 50 volts.

Voltage and Cross for resistance @ t = ∞

As we know the voltage drop across inductor reaches to its minimum when a long time has passed, which is given by, V_L(min) = 20 V, voltage drop across resistor is emf - V_L(min) = 50 - 20 V = 30 V.

At this point, using ohm’s law current through the resistor is given by, i = V/R = 3A

Therefore, the voltage drop across resistor is 30 volts and the current through resistor is 3 amperes.

Resistance of solenoid

Using ohm’s law to get the resistance of coil is

$R_L=\frac{V_L}{l}=\frac{20V}{3A}=6.7\Omega$

Therefore, the resistance of the solenoid is 6.7 ohms.

Self-inductance of the coil

By step 2, voltage drop across coil is given by

$V_L=i_0{R}_L=\frac{\epsilon R_L}{\left(R_0+R\right)}\left(1-e^{-\left(\frac{R_0+R}{L}\right)t}\right)$

Put all the value calculated in the equation, V_L = 31 volts.

From the graph, at V_L = 31 volts, time constant is 2.3 ms. Thus, self-inductance of the coil is given by

$\begin{array}{r}L=\tau \left(R+R_L\right)\\ =\left(2.3\mathrm{ms}\right)(10+6.7)\mathrm{\Omega }\\ =38.41\mathrm{mH}\end{array}$

Therefore, the self-inductance of the coil is 38.41 milli-henry.