A step-down transformer is connected to a 240 | vedclass

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(C) Given:Primary voltage $(V_p)$ = $2400 \, V$Secondary current $(I_s)$ = $80 \, A$Turns ratio $(N_p : N_s)$ = $20 : 1$Efficiency $(\eta)$ = $100\%$F

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$4$

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(C) Given:Primary voltage $(V_p)$ = $2400 \, V$Secondary current $(I_s)$ = $80 \, A$Turns ratio $(N_p : N_s)$ = $20 : 1$Efficiency $(\eta)$ = $100\%$For an ideal transformer, the ratio of voltages is equal to the ratio of turns:$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$\frac{V_s}{2400} = \frac{1}{20}$$V_s = \frac{2400}{20} = 120 \, V$Since the efficiency is $100\%$, the input power equals the output power:$P_{in} = P_{out}$$V_p 	imes I_p = V_s 	imes I_s$$2400 	imes I_p = 120 	imes 80$$2400 	imes I_p = 9600$$I_p = \frac{9600}{2400} = 4 \, A$Therefore, the current in the primary coil is $4 \, A$.

$A$ step-down transformer is connected to a $2400 \, V$ line, and $80 \, A$ of current is found to flow in the output load. The ratio of the turns in the primary and secondary coil is $20 : 1$. If the transformer efficiency is $100\%$, then the current flowing in the primary coil will be: (in $A$)

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