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Question
Describe briefly, with the help of labelled diagram, working of a step-up transformer.
A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.
Solution
In a transformer with Ns secondary turns and Npprimary turns, induced emf or voltage Es is:
`E_s = -N_s (dphi)/dt`
Back emf = Ep =`-N_p(dphi)/(dt)`
EP = VP
Es = Vs
Thus, Vs = `-N_s (dphi)/()dt`… (1)
`V_p =-N_p (dphi)/dt ... (2)`
Dividing equations (i) and (ii), we obtain
`(V_s)/(V_p) = N_s/N_p`
If the transformer is 100% efficient, then
`i_p v_p = i_s v_s` = Power (p)
Thus, combining the above equations,
`i_p/i_s = v_s/v_p = N_s/N_p`
Or,
`V_s = (N_s/N_p)V_p and I_s = (N_p/N_s) I_p`
If Ns > Np, then the transformer is said to be step-up transformer because the voltage is stepped up in the secondary coil.
No, the transformer does not violate the principal of conservation of energies. This can be easily observed by the following equation:
`V_s I_s =V_pI_p`
Power consumed in both the coils is the same as even if the voltage increases or current increases, their product at any instant remains the same.
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