Question

The input power of a device is 10 kW at 1000 V. The output power is 500 W while the output impedance is 20 ohms. According to this; determine the voltage gain in dB.

Answer 1

The voltage gain of the device is approximately -37 dB.

Step-by-step explanation:

To determine the voltage gain in dB, we can use the formula for power gain and convert it to dB scale. The power gain (G) is defined as the ratio of output power (P_out) to input power (P_in).

Power Gain (G) = P_out / P_in

Where:

P_out = Output Power = 500 W

P_in = Input Power = 10 kW = 10,000 W

Now, we need to find the voltage gain (A_v) from the power gain, considering that the output impedance (Z_out) is 20 ohms.

Voltage Gain (A_v) = √(Power Gain) * (Load Impedance / Source Impedance)

Since we want to find the gain in dB, we will use the logarithmic scale:

Gain (A_v, dB) = 20 * log10(A_v)

Now, let's calculate the voltage gain (A_v) first:

Power Gain (G) = P_out / P_in = 500 W / 10,000 W = 0.05

Voltage Gain (A_v) = √(0.05) * (20 ohms / 1000 ohms) = √(0.05) * 0.02 ≈ 0.0141

Now, calculate the gain in dB:

Gain (A_v, dB) = 20 * log10(0.0141) ≈ 20 * (-1.85) ≈ -37 dB

So, the voltage gain of the device is approximately -37 dB.