Applications of mutual induction
Applications of mutual induction1. The transformer- it converts an alternating voltage across one coil to a larger or smaller alternating voltage across the other. Since H.E.P is lost through transmission lines therefore it is stepped down before it being transmitted and stepped up again at the point of supply lines. In a step uptransformer the number of turns in the secondary coil (Ns) is higher than the number of turns in the primary coil (Np). In a step downtransformer the primary coil has more turns than the secondary coil. The relationship between the primary voltage and the secondary voltage is given by; Np / Ns = Vp / Vs.The efficiency of a transformer is the ratio of power in secondary coil (Ps) to power in primary coil (Pp), therefore efficiency = Ps /Pp × 100%. Step up transformer1. A current of 0.6 A is passed through a step up transformer with a primary coil of 200 turns and a current of 0.1 A is obtained in the secondary coil.Determine the number of turns in the secondary coil and the voltage across if the primary coil is connected to a 240 V mains.SolutionNp / Ns = Vp / Vs = Ip / Is = Ns =(0.6 × 200) / 0.1 = 1200 turnsVp = 240 V hence Vs = (240 ×1200) / 200 = 1440 V2. A step-up transformer has 10,000 turns in the secondary coil and 100 turns in the primary coil. An alternating current of 0.5 A flows in the primary circuit when connected to a 12.0 V a.c. supply.a) Calculate the voltage across the secondary coilb) If the transformer has an efficiency of 90%, what is the current in the secondary coil?Solutiona) Vs = (Ns / Np) × Vp = (10,000× 12) / 100 = 1200 Vb) Power in primary = Pp = Ip × Vp= 5.0 × 12 = 60 WEfficiency = Ps / Pp × 100% but Ps = Is VsIs = (60 × 90) / (1200 × 100) = 0.045 A3. Induction coil –was developed in 1851 by Heinrich Ruhmkortt. It has both secondary and primary coils with an adjustable spark gap.4. Car ignition system – it is applied in petrol driven engines where a spark plug is used to ignite petrol vapour and air mixture to run the engine.