If 1 watt of RF power is fed to a directional coupler having 30 dB coupling, the power available at the coupled port is
- A. $\frac{1}{30} W$
- B. $\frac{1}{10} W$
- C. $\frac{1}{100} W$
- D. $\frac{1}{1000} W$
Answer: D. $\frac{1}{1000} W$
Explanation: $Coupling (dB) = 10 \log_{10}(\frac{P_{in}}{P_{coupled}})$. With $P_{in} = 1$ W and Coupling = 30 dB, $30 = 10 \log_{10}(\frac{1}{P_{coupled}}) \Rightarrow 3 = \log_{10}(\frac{1}{P_{coupled}}) \Rightarrow \frac{1}{P_{coupled}} = 1000 \Rightarrow P_{coupled} = \frac{1}{1000}$ Watt.