The number of turns in the primary and second | vedclass

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(B) For an ideal transformer,the ratio of the primary voltage $(E_p)$ to the secondary voltage $(E_s)$ is equal to the ratio of the number of turns in

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$40$

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(B) For an ideal transformer,the ratio of the primary voltage $(E_p)$ to the secondary voltage $(E_s)$ is equal to the ratio of the number of turns in the primary coil $(N_p)$ to the number of turns in the secondary coil $(N_s)$.The transformer equation is given by: $\frac{E_p}{E_s} = \frac{N_p}{N_s}$Given values are:$N_p = 100$$N_s = 20$$E_p = 200 \, V$Substituting these values into the equation:$\frac{200}{E_s} = \frac{100}{20}$$\frac{200}{E_s} = 5$$E_s = \frac{200}{5} = 40 \, V$Therefore,the induced potential in the secondary coil is $40 \, V$.

The number of turns in the primary and secondary coils of a transformer are $100$ and $20$ respectively. If an alternating potential of $200 \, V$ is applied to the primary,the induced potential in the secondary will be......$V$.

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