a) Two identical rods are driven into the ground at A and B, introducing an unknown resistance \(\displaystyle{R}_{{y}}\). The procedure for finding the unknown resistance \(\displaystyle{R}_{{x}}\) is as follows: measure resistance \(\displaystyle{R}_{{1}}\) between points A and B. Then connect A and B with a heavy conducting wire, and measure resistance \(\displaystyle{R}_{{2}}\) between points A and C. This shown in below figure.

The equivalent resistance of series resistors \(\displaystyle{R}_{{y}}\) and \(\displaystyle{R}_{{y}}\) is

\(\displaystyle{R}_{{1}}={R}_{{y}}+{R}_{{y}}\)

\(\displaystyle={2}{R}_{{y}}\)

From the above circuit \(\displaystyle{R}_{{y}}\) and \(\displaystyle{R}_{{y}}\) are connected in parallel. The equivalent resistance is

\(\displaystyle{R}_{{{e}{q}}}={\left[{\frac{{{1}}}{{{R}_{{y}}}}}+{\frac{{{1}}}{{{R}_{{y}}}}}\right]}^{{-{1}}}\)

\(\displaystyle={\frac{{{R}_{{y}}}}{{{2}}}}\)

And, \(\displaystyle{R}_{{{e}{q}}}\) and \(\displaystyle{R}_{{x}}\) are connected in series. Then the equivalent resistance is

\(\displaystyle{R}_{{2}}={R}_{{{e}{q}}}+{R}_{{x}}\)

\(\displaystyle={\frac{{{R}_{{y}}}}{{{2}}}}+{R}_{{x}}\)

From (1),

\(\displaystyle{R}_{{y}}={\frac{{{R}_{{1}}}}{{{2}}}}\)

Substitute this value in equation, we get

\(\displaystyle\therefore{R}_{{2}}={\frac{{{1}}}{{{2}}}}{\left({\frac{{{R}_{{1}}}}{{{2}}}}\right)}+{R}_{{x}}\)

\(\displaystyle={\frac{{{R}_{{1}}}}{{{4}}}}+{R}_{{x}}\)

\(\displaystyle\therefore{R}_{{x}}={R}_{{2}}-{\frac{{{R}_{{1}}}}{{{4}}}}\)

b) If the measurement give \(\displaystyle{R}_{{1}}={13}\) Ohm and \(\displaystyle{R}_{{2}}={6.0}\) Ohm

Then

\(\displaystyle{R}_{{x}}={R}_{{2}}-{\frac{{{R}_{{1}}}}{{{4}}}}\)

\(\displaystyle={\left({6.0}\ {O}{h}{m}\right)}-{\left({\frac{{{13}\ {O}{h}{m}}}{{{4}}}}\right)}\)

\(\displaystyle={2.75}\) Ohm

\(\displaystyle\approx{2.8}\) Ohm

Here satisfactory ground resistance would be \(\displaystyle{R}_{{x}}{<}{2.0}\) Ohm. So \(\displaystyle{R}_{{x}}\) is exceeds the limit of 2.0 Ohm

Therefore, the antenna is not adequate grounded.