Final answer:
Without considering the effect of inductance on AC circuits, the current drawn by the coil when the supply frequency increases to 60 Hz will remain the same as at 50 Hz, which is 5 A. This is because the power consumption is given as constant, and thus, with constant resistance, the current remains the same according to the power equation P = I^2R.
Step-by-step explanation:
To calculate the value of the current drawn by the coil if the supply frequency increases to 60 Hz, we need to use the power equation for AC circuits and consider the properties of the coil as an inductor. The power consumed by an inductor is due to its resistance since ideal inductors do not consume power in the reactive part of the circuit. The power equation can be expressed as P = I2R, where P is power, I is current, and R is resistance.
Given that the power consumption is 600 W when the current is 5 A, we can calculate the resistance (R) of the coil using the formula:
R = P / I2
Since the power is constant and does not change with frequency, we can say that:
Inew = sqrt(P/R)
In this scenario, however, the question does not provide enough information to directly calculate the change in current with frequency, because the inductance of the coil plays a significant role at different frequencies. Normally, with increased frequency, the inductive reactance (XL) of the coil increases, which could affect the current. But because we were instructed to ignore the inductance, the current will remain dependent only on the resistance and the constant power consumption.
If the power consumed remains the same at the new frequency and the resistance does not change, then the current drawn by the coil at 60 Hz remains the same as at 50 Hz, which is 5 A.