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JIPMER 2009
Exam
JIPMER
Year
2009
Subject
Physics
Chapter
Alternating Current

JIPMER 2009 · Physics · Alternating Current

The primary and secondary coils of a transformer have 50 and 1500...

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+4 t, where ϕ\phi is in weber, tt is time in second and ϕ0\phi_0 is a constant, the output voltage across the secondary coil is

Options

  1. A90 V
  2. B120 V
  3. C220 V
  4. D30 V

Correct answer

B. 120 V

Step-by-step solution

The magnetic flux linked with the primary cc is given by ϕ=ϕ0+4t\phi=\phi_0+4 t So, voltage across primary Vp=dϕdt=ddt(ϕ0+4t)=4volt( as ϕ0= consta \begin{aligned} V_p & =\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right) \\ & =4 \mathrm{volt} \quad\left(\text{ as } \phi_0=\right.\text{ consta } \end{aligned} Also, we have Np=50 and  N_p=50 \text{\text{ and } } N_s=1500 From relation, or VsVp=NsNpVs=VpNsNp=4(150050)=120V\begin{aligned} \frac{V_s}{V_p} & =\frac{N_s}{N_p} \\ V_s & =V_p \frac{N_s}{N_p} \\ & =4\left(\frac{1500}{50}\right) \\ & =120 \mathrm{ V} \end{aligned}
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