91Ó°ÊÓ

Using the definition of efficiency, the efficiency of a class B push-pull amplifier can be expressed as: \[Efficiency = \frac{Output\ power}{Output\ power + Power\ dissipated}\] To calculate the output power, remember that a sinusoidal waveform is present at the output and that each transistor amplifies only one half-cycle of the sinusoidal signal. When the active transistor is fully turned on, the voltage across the load is maximum and equal to the supply voltage, and the current through the load is also maximum. Therefore, OUTPUT_POWER = VOLTAGE * CURRENT. After doing the necessary calculations, we get \[Output\ power = \frac{1}{\pi}V_{supply}I_{max}\] Where \(V_{supply}\) represents the supply voltage, and \(I_{max}\) represents the maximum current. For a class B push-pull amplifier, the power dissipated by the output transistors is minimal because they are either conducting (turned on) or not conducting (turned off). Therefore, we can assume that the power dissipation is negligible.

Now, we can plug the known values back into the efficiency equation: \[Efficiency = \frac{\frac{1}{\pi}V_{supply}I_{max}}{\frac{1}{\pi}V_{supply}I_{max} + Power\ dissipated}\] As we previously mentioned, the power dissipation for a class B push-pull amplifier is negligible, so we can eliminate it from the equation. We now have: \[Efficiency = \frac{\frac{1}{\pi}V_{supply}I_{max}}{\frac{1}{\pi}V_{supply}I_{max}}\] This simplifies to: \[Efficiency = 1\] The maximum theoretical efficiency of a class B push-pull amplifier is 100%, but in reality, the actual efficiency is slightly less than this number due to non-zero power dissipation. Therefore, the correct answer is (d) \(98 \%\).