Final answer:
The voltage applied to a series LR circuit with an 8V drop across the inductor and a 6V drop across the resistor is 10V, and the power factor of the circuit is 0.6.
Step-by-step explanation:
In a series LR circuit, the voltage drop across the inductor (L) and the resistor (R) are given as 8V and 6V, respectively. To find the total voltage applied to the circuit (V), we use the Pythagorean theorem because the resistor voltage and inductor voltage are perpendicular to each other in a phasor diagram, due to the phase difference caused by the inductor. Hence, the voltage applied (V) is given by √(82 + 62) = √(64 + 36) = √100 = 10V.
The power factor (cos φ) of the circuit is the cosine of the phase angle between the voltage across the entire LR circuit and the current through it. It can be calculated as the resistance (R) divided by the impedance (Z). Since we have a right triangle with the adjacent side (voltage drop across resistor) as 6V and the hypotenuse (applied voltage) as 10V, the power factor is 6V/10V = 0.6.
Therefore, the voltage applied to the circuit is 10V and the power factor is 0.6. The correct answer is C. 10V,0.6.