An ultrasound system is set to 0db and is transmitting at full intensity. What is the output power when the system is transmitting at 50% of full intensity?

• A. -3 dB
• B. -50 dB
• C. 3 dB
• D. -10 dB

Answer: (A)

The key idea is how decibels relate to power. A change in power expressed in dB is 10 times the log base 10 of the ratio P2/P1. If you go from full power to 50% of that power, the ratio is 0.5. So the change is 10 log10(0.5) ≈ -3.01 dB. Since the system is at 0 dB for full power, transmitting at 50% yields about -3 dB output relative to that full-power reference.

This matches the option that indicates a small reduction of roughly 3 dB. The other options represent much larger or different changes in power: -50 dB would be an enormous drop (to one hundred-thousandth of the power), -10 dB is a tenfold reduction, and +3 dB would actually double the power.