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(A) For an ideal transformer,the power input equals the power output: $P_p = P_s$.Since $P = V \cdot i$,we have $V_p \cdot i_p = V_s \cdot i_s$,which

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$4 : 5$

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(A) For an ideal transformer,the power input equals the power output: $P_p = P_s$.Since $P = V \cdot i$,we have $V_p \cdot i_p = V_s \cdot i_s$,which implies $\frac{i_p}{i_s} = \frac{V_s}{V_p}$.Also,for a transformer,the ratio of voltages is equal to the ratio of the number of turns: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.Combining these,we get the relationship: $\frac{i_p}{i_s} = \frac{N_s}{N_p}$.Given the number of turns in the primary coil $N_p = 5$ and the secondary coil $N_s = 4$,the ratio of the currents is $\frac{i_p}{i_s} = \frac{4}{5}$ or $4 : 5$.

In a transformer,the number of turns in the primary coil and secondary coil are $5$ and $4$ respectively. If $240 \ V$ is applied to the primary coil,then the ratio of the current in the primary coil to the secondary coil is:

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