Alternating Current Ques 32
32. The primary and secondary coil of a transformer have $50$ and $1500$ turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi=\phi_0+4 t$, where $\phi$ is in webers, $t$ is time in seconds and $\phi_0$ is a constant, the output voltage across the secondary coil is
[2007]
(a) $120$ volts
(b) $220$ volts
(c) $30$ volts
(d) $90$ volts
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Answer:
Correct Answer: 32.(a)
Solution:
- (a) Since $\frac{V_s}{V_p}=\frac{N_s}{N_p}$
Whereas
$N_s=$ No. of turns across secondary coil $=50$
$N_p=$ No. of turns in secondary coil $=1500$
and $V_p=\frac{d \phi}{d t}=\frac{d}{d t}(\phi_0+4 t)=4$
$\Rightarrow V_s=\frac{1500}{50} \times 4=120 $ $V$
BETA
