Correct Option: (A) 10 V
Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
In an L-R series circuit connected to an AC source, the voltage across the resistor ($V_R$) and the voltage across the inductor ($V_L$) are not in phase. The total source voltage ($V$) is the phasor sum of these voltages, given by the formula:
$$V = \sqrt{V_R^2 + V_L^2}$$Given data:
- Voltage across resistor, $V_R = 8\text{ V}$
- Voltage across inductor, $V_L = 6\text{ V}$
Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
Substituting the given values into the phasor sum formula:
$$V = \sqrt{(8)^2 + (6)^2}$$ $$V = \sqrt{64 + 36}$$ $$V = \sqrt{100}$$Step 3 (Final Concluded Answer): [0.25 Mark]
Calculating the square root, we get:
$$V = 10\text{ V}$$Thus, the voltage of the source is 10 V, which corresponds to option (A).
💡 EL10 Concept Insight: Because the inductor makes the current "lag" behind the voltage, the voltages don't line up perfectly to just add them together; instead, they act like the sides of a right-angled triangle, so we use the Pythagorean theorem to find the total "hypotenuse" voltage.