The voltage ratio in a transformer is equal to the turns ratio:
\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
where \( V_s \) and \( V_p \) are the secondary and primary voltages, and \( N_s \) and \( N_p \) are the number of turns in the secondary and primary coils, respectively. Given \( V_p = 12 \, \text{kV} = 12000 \, \text{V} \) and \( V_s = 120 \, \text{V} \), we have:
\( \frac{120}{12000} = \frac{N_s}{N_p} \quad \Rightarrow \quad \frac{N_s}{N_p} = \frac{1}{100} \)
For an ideal transformer, the input power is equal to the output power:
\( P_{in} = P_{out} \)
Power is given by \( P = IV \), so:
\( I_p V_p = I_s V_s = 60 \, \text{kW} = 60000 \, \text{W} \)
We can calculate the primary current (\( I_p \)):
\( I_p = \frac{60000}{V_p} = \frac{60000}{12000} = 5 \, \text{A} \)
Similarly, we can calculate the secondary current (\( I_s \)):
\( I_s = \frac{60000}{V_s} = \frac{60000}{120} = 500 \, \text{A} \)
The secondary resistance (\( R_s \)) can be calculated using Ohm’s law:
\( R_s = \frac{V_s}{I_s} = \frac{120}{500} = 0.24 \, \Omega = 240 \, \text{m}\Omega \)
The value of the resistive load in the secondary circuit is \( 240 \, \text{m}\Omega \).