A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5 dB per km and cable length is 20 km. The power received at receiver is `10^(-x)` W. The value of x is _____ . [Gain in `10 log_(10) ((P_0)/(P_i))` ]

Answer

To solve the problem, we need to calculate the power received at the receiver after considering the attenuation over the length of the cable. Here are the steps to find the value of \( x \):

### Step 1: Understand the given values
- Transmitted power \( P_0 = 0.1 \, \text{kW} = 0.1 \times 10^3 \, \text{W} = 100 \, \text{W} \)
- Attenuation of the cable = -5 dB/km
- Length of the cable = 20 km

### Step 2: Calculate total attenuation
The total attenuation over the length of the cable can be calculated as:
\[ \text{Total Attenuation (in dB)} = \text{Attenuation per km} \times \text{Length of cable} \]
\[ \text{Total Attenuation} = -5 \, \text{dB/km} \times 20 \, \text{km} = -100 \, \text{dB} \]

### Step 3: Convert attenuation to power ratio
The relationship between power and decibels is given by:
\[ \text{Gain (in dB)} = 10 \log_{10} \left( \frac{P_0}{P_i} \right) \]
Since we have a loss, we can express it as:
\[ -100 = 10 \log_{10} \left( \frac{P_0}{P_i} \right) \]

### Step 4: Solve for the power ratio
Dividing both sides by 10:
\[ -10 = \log_{10} \left( \frac{P_0}{P_i} \right) \]
Taking the antilogarithm:
\[ \frac{P_0}{P_i} = 10^{-10} \]
Thus, we can express \( P_i \) in terms of \( P_0 \):
\[ P_i = \frac{P_0}{10^{10}} \]

### Step 5: Substitute the value of \( P_0 \)
Substituting \( P_0 = 100 \, \text{W} \):
\[ P_i = \frac{100}{10^{10}} = 10^{2} \times 10^{-10} = 10^{-8} \, \text{W} \]

### Step 6: Compare with the given expression
According to the problem, the power received at the receiver is given as:
\[ P_i = 10^{-x} \, \text{W} \]
From our calculation, we found:
\[ P_i = 10^{-8} \, \text{W} \]
Thus, comparing both expressions:
\[ -x = -8 \implies x = 8 \]

### Final Answer
The value of \( x \) is \( \boxed{8} \).

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