Final answer:
To find the current in the circuit with an inductive load and non-inductive resistor in series, the voltage readings across both elements are used with the Pythagorean theorem to obtain the total rms voltage across the series combination. Applying Ohm's law to this voltage and the given resistance of 8 ohms, the current in the circuit is calculated to be 10 A.
Step-by-step explanation:
The question asks us to calculate the current in the circuit where an inductive load and a non-inductive resistor are connected in series across an AC supply. The voltmeter readings across the resistor and the inductive load are 64 V and 48 V respectively. According to the principle of voltages in series, the total voltage supplied is the vector sum of the individual voltages across the resistor and the inductor. Therefore, we can apply the Pythagorean theorem because the resistor voltage (64 V) is in phase with the current, while the inductor voltage (48 V) leads the current by 90 degrees in an inductive circuit.
Using the Pythagorean theorem: Total voltage (¹°°V) = √(Resistor voltage)2 + (Inductor voltage)2
= √(64V)2 + (48V)2 = 80 V
Therefore, this is the rms voltage across the series combination of the resistor and the inductor.
Since the resistor is 8Ω and the total rms voltage across it and the inductor is 80V, the current can be found using Ohm's law (I = V/R), yielding:
Current (I) = Total voltage / Resistance
= 80 V / 8Ω = 10 A
Thus, the current in the circuit is 10 A.