F6-2: Mutual Inductance - Transformers

NB: Students are advised to perform experiment F6-1 Self Inductance before attempting F6-2.

Apparatus

Coils 300 turns & 150 turns; wire for 20 turn & 10 turn coils; 2 iron C-cores; C-core clip; \(10\Omega\) and \(5\Omega\) resistors; CRO (oscilloscope); AC power supply; connecting leads (5 short).

NB: This experiment requires mains electricity.

Procedure

  1. Construct the following:

    F6-2.1

  1. Using \(N_p\) = 150 turns, \(N_s\) = 10 turns. Measure and set \(V_{in} = V_p = 4 \text{V}\) peak. Measure \(V_s\) peak. Calculate \(\frac{V_p}{V_s}\) and \(\frac{N_p}{N_s}\) and compare their values.

  2. Repeat a) with the following numbers of turns:

    F6-2.2

  3. Keeping \(N_p = 150\) turns and \(N_s = 300\) turns, reduce \(V_{in} = V_p\) to \(2\text{V}\) peak. Measure \(V_s\) peak, and calculate and compare \(\frac{V_p}{V_s}\) and \(\frac{N_p}{N_s}\).

  4. Remove the clip from the iron cores, and remove one C-core. Place two coils, one on each arm of a single C-core. Again use \(N_p = 150\) turns and \(N_s = 300\) turns. Set \(V_{in} = V_p = 4\text{V}\) peak. Measure \(V_s\).

  1. Construct the following with \(N_p = 300\) turns, \(N_s = 150\) turns, and an extra test coil of 10 turns as shown:

    F6-2.3

  1. Connect the CRO to measure \(V_p\) peak, and set \(V_{in}\) so that \(V_p = 4\)V peak.
  2. Measure \(V_R\) peak, and thus calculate \(I_p\) peak, the current in the primary circuit.
  3. Measure \(V_T\) peak, across the 10 turn test coil.
  4. Connect a \(5\Omega\) resistor between A and B, across the 150 turn secondary coil. Connect the CRO to measure \(V_p\) and adjust \(V_{in}\) so that \(V_p = 4\)V peak.
  5. Measure \(V_R\) peak, and thus calculate \(I_p\) peak.
  6. Measure \(V_T\) peak again. This should be about the same size as the value measured in procedure 2 c) above.

Theory

  1. Flux \(\Phi\) and induced emf \(E\) are related by the following:

    \[E_p = -N_p \frac{d \Phi_s}{dt} \label{eqnA} \tag{equation A}\]
    \[E_s = -N_s \frac{d \Phi_p}{dt} \label{eqnB} \tag{equation B}\]

    The unit of flux is the weber - Wb. When \(\Phi_s\) (the flux through the secondary coil) \(=\Phi_p\) (the flux through the primary coil), then:

    \[\frac{d \Phi_s}{dt} = \frac{d \Phi_p}{dt}\]

    therefore:

    \[\frac{E_s}{N_s} = \frac{E_p}{N_p}\]

    thus:

    \[\frac{E_p}{E_s} = \frac{N_p}{N_s}\]

    If the resistances of the primary coil and the secondary coil are both low and the currents flowing through them are not too large, then:

    \[V_p \approx E_p \quad \text{and} \quad V_s \approx E_s\]

  2. The 10 turn coil is used to detect if the flux \(\Phi\) in the iron core changes in the experiment. If \(\Phi = \Phi_{peak} \sin \omega t\), then \(\ref{eqnB}\) can be used to show that \(V_T\) peak \(\propto \Phi_{peak}\), provided that \(\omega\) is constant.

Analysis

  1. Why, in experiment 1a) to 1c), are the two calculated ratios not exactly equal (hint: use the theory, and the fact that the primary coil has some resistance)?
  2. Use the theory to explain the result of procedure 1 d).
  3. According to Lenz’s Law, the induced current in the secondary coil in the procedure 2 d) is in such a direction so as to reduce the flux in the core. However procedure 2 f) shows that the flux remains approximately constant. How is this possible (hint: consider the primary coil)?
  4. Give an explanation in terms of power flow for the change in \(I_p\) produced as a result of connecting the \(5\Omega\) resistor to the secondary coil.
    1. These coils are simple electrical transformers. What are the causes of power loss in a transformer, and how can they be minimised?
    2. All electricity supply companies use transformers in their power distribution systems. Explain, giving reasons, how they are used.