If only 5% of the supplied power to a cable reaches the output terminal, the power loss in the cable, in decibels, is _______. (round off to nearest integer)

The correct answer is 13.

Step-by-Step Explanation:

We are given that only 5% of the supplied power reaches the output terminal of the cable. Let us define the input (supplied) power as $P_{\text{in}}$ and the output power as $P_{\text{out}}$.

According to the problem statement:
$P_{\text{out}}=5\%\text{of}{P}_{\text{in}}=0.05\times P_{\text{in}}$

Power loss or attenuation in decibels (dB) represents the ratio of the input power to the output power, expressed on a logarithmic scale. The formula for the power loss in decibels is:
$\text{Power Loss (dB)}=10{\log }_{10}\left(\frac{P_{\text{in}}}{P_{\text{out}}}\right)$

Substituting the expression for $P_{\text{out}}$ into the formula:
$\text{Power Loss (dB)}=10{\log }_{10}\left(\frac{P_{\text{in}}}{0.05\times P_{\text{in}}}\right)$

The term $P_{\text{in}}$ cancels out:
$\text{Power Loss (dB)}=10{\log }_{10}\left(\frac{1}{0.05}\right)$

Since $\frac{1}{0.05}=20$, we have:
$\text{Power Loss (dB)}=10{\log }_{10}\left(20\right)$

Using the properties of logarithms, we know that:
${\log }_{10}\left(20\right)={\log }_{10}\left(2\times 10\right)={\log }_{10}\left(2\right)+{\log }_{10}\left(10\right)$

Substituting the values ${\log }_{10}\left(2\right)\approx 0.3010$ and ${\log }_{10}\left(10\right)=1$:
${\log }_{10}\left(20\right)\approx 0.3010+1=1.3010$

Now, calculate the final value in decibels:
$\text{Power Loss (dB)}=10\times 1.3010=13.010\text{dB}$

Rounding to the nearest integer, the power loss in the cable is 13 dB.