A transformer has 100 turns in the primary co | vedclass

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(C) Given: Primary turns $N_p = 100$, Primary current $I_p = 8 \, A$, Input power $P_{in} = 1000 \, W$, Secondary voltage $V_s = 500 \, V$. First, cal

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$400$

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(C) Given: Primary turns $N_p = 100$, Primary current $I_p = 8 \, A$, Input power $P_{in} = 1000 \, W$, Secondary voltage $V_s = 500 \, V$. First, calculate the primary voltage $V_p$ using $P_{in} = V_p I_p$: $V_p = \frac{P_{in}}{I_p} = \frac{1000 \, W}{8 \, A} = 125 \, V$. Using the transformer turns ratio formula: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. Substituting the values: $\frac{125}{500} = \frac{100}{N_s}$. $\frac{1}{4} = \frac{100}{N_s} \Rightarrow N_s = 400$.

$A$ transformer has $100$ turns in the primary coil and carries $8 \, A$ current. If the input power is $1 \, kW$, the number of turns required in the secondary coil to have $500 \, V$ output is:

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