# F6-1: Self Inductance in AC Circuits¶

## Apparatus¶

300 turn coil; 2 iron C-cores; C-core clip; ammeter $$(0 - 1\text{Adc})$$; voltmeter $$(0 - 5\text{Vdc})$$; 3 batteries $$(3\text{V})$$; 2 $$\times 100\Omega$$ resistors; CRO (oscilloscope); AC power supply; $$0.5\text{m}$$ ruler; connecting leads (5 short); 1 sheet graph paper.

NB: This experiment requires mains electricity.

## Procedure¶

1. To find the resistance $$r$$ of the coil, connect the following circuit and use the readings of the ammeter and voltmeter to calculate $$r$$:

1. To find the inductance $$L$$ of the coil with an iron core, connect the following circuit:

1. Connect the CRO input to $$V_1$$. Adjust the CRO so that $$V_1$$ is about $$4\text{V}$$, the supply being switched to AC. Adjust the CRO so that the AC waveform is seen clearly:

Carefully adjust the power supply so that the peak value of $$V_1$$ as seen on the screen is $$4\text{V}$$ peak.

1. Now connect the CRO to measure $$V_R$$ peak. Use this, and the value of R to calculate the value of $$I$$ peak for the circuit.
2. Use the CRO to measure $$V_2$$ peak.
3. Repeat a), b), and c), but using $$V_1$$ of $$3.5, 3, 2.5, 2, 1.5, 1, 0.5, \text{ and } 0\text{V}$$ peak each time.
1. Tabulate the values of $$V_R$$ peak, $$V_2$$ peak, and $$I$$ peak.
2. Measure the cross-sectional area of the iron core and the distance $$l$$ around the centre of the pair of cores as shown below:

## Theory¶

The coil with its iron core has an inductance $$L$$, and the copper wire of the coil has a resistance $$r$$. An equivalent circuit for the coil is:

The impedance of this combination of $$r$$ and $$L$$ is given by:

$Z = \sqrt{r^2 + \omega ^2 L^2} \label{eqn1} \tag{equation 1}$

where:

$\qquad Z = \frac{V_2 \text{ rms}}{I \text{ rms}} = \frac{V_2 \text{ peak}}{I \text{ peak}} \label{eqn2} \tag{equation 2}$

and:

$\qquad \omega = 2 \pi f$

It is also possible to calculate $$L$$ using data about the coil and its core:

$L = \frac{ \mu_0 \mu_r N^2 A}{l} \label{eqn3} \tag{equation 3}$

where:

$\begin{split}\mu_0 &= 4 \pi \times 10^{-7} \text{Hm} ^{-1} \\ \mu_r &= \text{relative permeability of the iron} \\ N &= \text{number of coil turns} \\ A &= \text{core cross-sectional area} \\ l &= \text{distance around core}\end{split}$

## Analysis¶

1. Plot a graph of $$V_2$$ peak against $$I$$ peak, and find the gradient.
2. Use the gradient and $$\ref{eqn2}$$ to find the value of $$Z$$.
3. Use $$\ref{eqn1}$$ to find the inductance of the coil, $$L$$.
4. Use this value of $$L$$, together with the other measured values, and $$\ref{eqn3}$$, to find the relative permeability of this type of iron, $$\mu_r$$.
5. Look up values of $$\mu_r$$ for different types of iron in a reference table and try to deduce the type of iron alloy used in your iron cores.

## Questions¶

1. Use the values of $$R$$, $$r$$, and $$L$$ above to calculate the current $$I$$ peak when $$V_1 = 8$$V peak and $$f = 1$$kHz.

2. Use $$\ref{eqn3}$$ to estimate $$L$$ if the iron core is removed. Show that in this case $$Z \approx r$$. Repeat Q1 using $$L$$ for the coil without an iron core.

3. Briefly explain the energy changes in $$r$$ and $$L$$ when:

1. The current is + and rising
2. The current is a maximum +, and constant
3. the current is + and falling
4. How can an inductor be constructed so that power losses are kept to a minimum (consider both the design of the coil and the core)?

5. Why is it desirable to keep power losses to a minimum in an inductor used in the tuning circuit of a radio receiver?

6. Prove that:

$\qquad \frac {V_2 \text{ rms}}{I \text{ rms}} = \frac{V_2 \text{ peak}}{I \text{ peak}} \tag{from equation 2}$

Why in the experiment is $$V$$ peak measured rather than $$V$$ rms?

7. Sketch on the same graph curves of $$V$$ and $$I$$ for an inductor (with $$r = 0$$) which is connected to an AC power supply.

8. Explain carefully why $$V_1 \text{ peak} \neq V_R \text{ peak} + V_2 \text{ peak.}$$