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NEET PHYSICSEasy

The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux ϕ\phi linked with the primary coil is given by ϕ=ϕ0+4t\phi = \phi_0 + 4t, where ϕ\phi is in Weber, tt is time in seconds, and ϕ0\phi_0 is a constant, the output voltage across the secondary coil is:

A

90 V

B

120 V

C

220 V

D

30 V

Step-by-Step Solution

  1. Induced EMF in Primary (VpV_p): According to Faraday's law of induction, the induced electromotive force (emf) is the rate of change of magnetic flux. Given the flux linked with the primary coil is ϕ=ϕ0+4t\phi = \phi_0 + 4t, the induced voltage in the primary is: Vp=dϕdt=ddt(ϕ0+4t)=4 VV_p = \left| \frac{d\phi}{dt} \right| = \frac{d}{dt}(\phi_0 + 4t) = 4 \text{ V}

  2. Transformer Equation: For an ideal transformer, the ratio of voltages across the primary and secondary coils is equal to the ratio of their number of turns: VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p} Where: VsV_s is the secondary voltage (output). Vp=4 VV_p = 4 \text{ V} (primary voltage). Ns=1500N_s = 1500 (secondary turns). Np=50N_p = 50 (primary turns).

  3. Calculation: Vs=Vp×NsNpV_s = V_p \times \frac{N_s}{N_p} Vs=4×150050V_s = 4 \times \frac{1500}{50} Vs=4×30V_s = 4 \times 30 Vs=120 VV_s = 120 \text{ V}

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