Explanation: For a Class A audio amplifier, proper operation requires that the control grid never goes positive with respect to the cathode (assuming the cathode is at ground potential). If the grid goes positive, it will draw current, leading to distortion and loading of the preceding stage. With a grid bias of -10V, the instantaneous voltage on the grid is the sum of the DC bias and the instantaneous AC audio signal voltage. To prevent the grid from going positive, the positive peak of the AC audio signal cannot exceed the absolute value of the bias voltage. Therefore, the maximum permissible peak AC audio voltage is 10V. The question asks for the maximum permissible RMS value. For a sine wave, the RMS value is the peak value divided by the square root of 2 (approximately 1.414). RMS voltage = Peak voltage / $\sqrt{2}$ RMS voltage = 10 V / $\sqrt{2}$ $\approx$ 10 V / 1.414 = 7.07 V Thus, the maximum permissible RMS value is 7.07 V. * **A) 7.07 V:** This is correct, as calculated from the maximum peak voltage divided by $\sqrt{2}$. * **B) 8 V:** If the RMS voltage were 8V, the peak voltage would be $8 \text{ V} \times \sqrt{2} \approx 11.3 \text{ V}$. This would drive the grid positive by approximately 1.3V, causing distortion and grid current. * **C) 10 V:** This is the maximum *peak* voltage that can be applied to the grid, not the RMS voltage. If 10V RMS were applied, the peak voltage would be $10 \text{ V} \times \sqrt{2} \approx 14.14 \text{ V}$, which would drive the grid positive by 4.14V. * **D) 14.14 V:** This value, if interpreted as RMS, would result in a peak voltage of approximately $14.14 \text{ V} \times \sqrt{2} \approx 20 \text{ V}$. This would drive the grid 10V positive, causing severe distortion and heavy grid current.