Experimental determination of mutual inductance

The circuit arrangement for the measurement of mutual inductance between two coils P and S. C is a four-segment commutator and r is a very small resistance of the order of 0.01 ohm.

At first, 1 and 2 are connected together so that the secondary circuit is closed through the ballistic galvanometer (B.G.). Now segments 3 and 4 are also connected together to short circuit the resistance r.

When the key K is presented, the B.G. gives a throw. On pressing the key K, the current in the primary slowly grows. Hence an induced emf is produced in the secondary. Let l be the instantaneous current in the primary.

The emf induced in the secondary} = ε = - M dI /dt
The instantaneous current I, in the secondary is I’ = ε /R = M/R dI /dt, (numerically)

Where R is the total resistance of the secondary circuit.
Hence the total charge passing through the B.G., as the current in the primary grows from zero to a steady maximum value I0 in the time interval, t, is
        q = ∫^t₀ I’ dt = ∫t0 M/R . dI /dt dt = ∫^t₀ M/R dl = MI₀ /R            … (1)

If θ₁ is the first throw in the B.G., due to this charge, then
        q = T/2π . C/NBA .θ₁ [1 + λ/2]

    . :    M/R I₀ = T/2π C/NBA θ₁ [1 + λ/2]                    … (2)

To eliminate I₀ and C /(NBA) from Eq. (2), the contact between 1 and 2, and that between 3 and 4 are broken. The contact between 1 and 2, and that between 2 and 4 are made. The resistance r is now included in the primary circuit. As the value of r is very small, the steady current I0 in the primary circuit I0 in the primary circuit is not altered. The potential difference across r is I0r. It sends a steady current I0r/R through the B.G.

If Φ be the steady deflection corresponding to this current, then
    I_0r/R = C /NBA Φ                             … (3)
Dividing Eq. (2) by Eq. (3), we get
    M = rT/2π θ₁/Φ [1 + λ/2]

Knowing the time period T and the logarithmic decrement λ of the ballistic galvanometer, M can be calculated.