In a step-up transformer, how does the current in the secondary coil compare to the primary coil, assuming ideal conditi
**Transformer** works on mutual induction, V_s/V_p = N_s/N_p, I_s/I_p = N_p/N_s for ideal (power conserved V_p I_p = V_s I_s), step-up N_s>N_p V_s>V_p I_s V_p and N_s > N_p ), power is conserved ( V_p I_p = V_s I_s ). Since the secondary voltage is higher, the secondary current must be lower than the primary current ( I_s = I_p × (N_p/N_s) ), where (N_p/N_s) < 1 . Applying X_L = ωL, X_C = 1/ωC, Z =
Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations