The coils of a step-down transformer have 500 | vedclass

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(C) For an ideal transformer,the relationship between turns,voltage,and current is given by: $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{i_s}{i_p}$.Giv

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$40 \, A, 220 \, V$

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(C) For an ideal transformer,the relationship between turns,voltage,and current is given by: $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{i_s}{i_p}$.Given that it is a step-down transformer,the primary coil must have more turns than the secondary coil. Thus,$N_p = 5000$ and $N_s = 500$.Given values: $V_p = 2200 \, V$,$i_p = 4 \, A$.Using the ratio: $\frac{5000}{500} = \frac{2200}{V_s} = \frac{i_s}{4}$.Calculating voltage $V_s$: $\frac{5000}{500} = \frac{2200}{V_s} \Rightarrow 10 = \frac{2200}{V_s} \Rightarrow V_s = 220 \, V$.Calculating current $i_s$: $\frac{5000}{500} = \frac{i_s}{4} \Rightarrow 10 = \frac{i_s}{4} \Rightarrow i_s = 40 \, A$.Therefore,the current is $40 \, A$ and the potential difference is $220 \, V$.

The coils of a step-down transformer have $500$ and $5000$ turns. In the primary coil,an $AC$ of $4 \, A$ at $2200 \, V$ is sent. The value of the current and potential difference in the secondary coil will be:

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