Q132 / 200 · NEET 2007 · Physics

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

  • Correct answer
    A
    120 volts
  • B
    220 volts
  • C
    30 volts
  • D
    90 volts.

Solution

The magnetic flux linked with the primary coil is given by \(\phi=\phi_0+4 t\) So, voltage across primary \(\begin{aligned} & V_p=\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right) \\ & \left.=4 \text { volt (as } \phi_0=\text { constant }\right) \end{aligned}\) Also, we have \(N_p=50 \text { and } N_s=1500\) From relation, \(\begin{aligned} & \frac{V_s}{V_p}=\frac{N_s}{N_p} \\ & \text {or } V_s=V_p \frac{N_s}{N_p} \\ & =4\left(\frac{1500}{50}\right) \\ & =120 \mathrm{~V} \end{aligned}\)

Hence, the correct answer is option: 'A'.

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