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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:
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Verified Answer
The correct answer is:
120 volts
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,
\(\begin{aligned}
& \frac{V_s}{V_p}=\frac{N_s}{N_p} \\
& \text {or } V_s=V_p \frac{N_s}{N_p} \\
& =4\left(\frac{1500}{50}\right) \\
& =120 \mathrm{~V}
\end{aligned}\)
\(\phi=\phi_0+4 t\)
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,
\(\begin{aligned}
& \frac{V_s}{V_p}=\frac{N_s}{N_p} \\
& \text {or } V_s=V_p \frac{N_s}{N_p} \\
& =4\left(\frac{1500}{50}\right) \\
& =120 \mathrm{~V}
\end{aligned}\)
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