An A.C. of $50$ Hz and $1$ A peak value flows in primary coil transformer whose mutual inductance is $1.5$ H. Then peak value of induced emf in the secondary coil is:

150

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150 π

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300

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200

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Solution

The correct option is B $300$ V
Time period of AC $T = \frac{1}{n} = \frac{1}{50} s$
$Δ t = \frac{T}{4} = \frac{1}{50} \times \frac{1}{4} = \frac{1}{200} s$
$Δ I = 0 - 1 = - 1 A$
$e = - M (\frac{Δ I}{Δ t}) = 1.5 \times \frac{- 1}{\frac{1}{200}} = - 300 V$

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An A.C. of 50 Hz and 1 A peak value flows in primary coil transformer whose mutual inductance is 1.5H. Then peak value of induced emf in the secondary coil is:

The correct option is B 300 V Time period of AC T=1n=150sΔt=T4=150×14=1200sΔI=0−1=−1A e=−M(ΔIΔt)=1.5×−11200=−300V

An A.C. of $50$ Hz and $1$ A peak value flows in primary coil transformer whose mutual inductance is $1.5$ H. Then peak value of induced emf in the secondary coil is:

The correct option is B $300$ V
Time period of AC $T = \frac{1}{n} = \frac{1}{50} s$
$Δ t = \frac{T}{4} = \frac{1}{50} \times \frac{1}{4} = \frac{1}{200} s$
$Δ I = 0 - 1 = - 1 A$
$e = - M (\frac{Δ I}{Δ t}) = 1.5 \times \frac{- 1}{\frac{1}{200}} = - 300 V$