For a periodic wave form, define the following: (i) Frequency (ii) period and (iii) Root-mean-square value

From the diagram, we identify a Delta network formed by resistors $R_{CF}=2\Omega$, $R_{FE}=3\Omega$, and $R_{EC}=4\Omega$.
The sum of these resistors is $R_{sum} = R_{CF} + R_{FE} + R_{EC} = 2\Omega + 3\Omega + 4\Omega = 9\Omega$.
We convert this Delta to an equivalent Star network with resistors $R_C'$, $R_F'$, $R_E'$ connected to a central node N.
Using the Delta-to-Star conversion formulas:

• Resistor from C to N ($R_C'$): $$R_C' = \frac{R_{CF} \times R_{EC}}{R_{sum}} = \frac{2\Omega \times 4\Omega}{9\Omega} = \frac{8}{9}\Omega$$
• Resistor from F to N ($R_F'$): $$R_F' = \frac{R_{CF} \times R_{FE}}{R_{sum}} = \frac{2\Omega \times 3\Omega}{9\Omega} = \frac{6}{9}\Omega = \frac{2}{3}\Omega$$
• Resistor from E to N ($R_E'$): $$R_E' = \frac{R_{FE} \times R_{EC}}{R_{sum}} = \frac{3\Omega \times 4\Omega}{9\Omega} = \frac{12}{9}\Omega = \frac{4}{3}\Omega$$

After the Delta-to-Star conversion, the original nodes C, F, E are now connected to the central node N via $R_C'$, $R_F'$, $R_E'$.
The circuit now has the following paths connected to node D:

• Path from N to D via F: $R_F'$ is in series with $R_{FD}=2\Omega$. $$R_{NFD} = R_F' + R_{FD} = \frac{2}{3}\Omega + 2\Omega = \frac{2+6}{3}\Omega = \frac{8}{3}\Omega$$
• Path from N to D via E: $R_E'$ is in series with $R_{ED}=2\Omega$. $$R_{NED} = R_E' + R_{ED} = \frac{4}{3}\Omega + 2\Omega = \frac{4+6}{3}\Omega = \frac{10}{3}\Omega$$

These two paths ($R_{NFD}$ and $R_{NED}$) are in parallel, connected between node N and node D. The equivalent resistance between N and D is:

$$R_{ND} = R_{NFD} \parallel R_{NED} = \frac{\frac{8}{3}\Omega \times \frac{10}{3}\Omega}{\frac{8}{3}\Omega + \frac{10}{3}\Omega} = \frac{\frac{80}{9}\Omega}{\frac{18}{3}\Omega} = \frac{\frac{80}{9}}{\frac{54}{9}}\Omega = \frac{80}{54}\Omega = \frac{40}{27}\Omega$$

The circuit is now a series combination of $R_{AC}$, $R_C'$, $R_{ND}$, and $R_{DB}$.
From the diagram:

• $R_{AC} = 2.17\Omega$
• $R_C' = \frac{8}{9}\Omega$
• $R_{ND} = \frac{40}{27}\Omega$
• $R_{DB} = 4\Omega$ (resistor between D and B)

The total equivalent resistance $R_{AB}$ is the sum of these series resistances:

$$R_{AB} = R_{AC} + R_C' + R_{ND} + R_{DB}$$ $$R_{AB} = 2.17\Omega + \frac{8}{9}\Omega + \frac{40}{27}\Omega + 4\Omega$$

To sum these, we convert to fractions with a common denominator (2700 for 100 and 27):

$$R_{AB} = \frac{217}{100}\Omega + \frac{8 \times 3}{27}\Omega + \frac{40}{27}\Omega + \frac{4}{1}\Omega$$ $$R_{AB} = \frac{217}{100}\Omega + \frac{24}{27}\Omega + \frac{40}{27}\Omega + 4\Omega$$ $$R_{AB} = \frac{217}{100}\Omega + \frac{64}{27}\Omega + 4\Omega$$ $$R_{AB} = \frac{217 \times 27}{2700}\Omega + \frac{64 \times 100}{2700}\Omega + \frac{4 \times 2700}{2700}\Omega$$ $$R_{AB} = \frac{5859 + 6400 + 10800}{2700}\Omega$$ $$R_{AB} = \frac{23059}{2700}\Omega$$ $$R_{AB} \approx \boxed{8.54\Omega}$$

2a) For a periodic wave form, define the following:

(i) Frequency

• Frequency ($f$) is the number of complete cycles or oscillations of a periodic waveform that occur per unit of time. It is typically measured in Hertz (Hz), where 1 Hz equals one cycle per second.

(ii) Period

• Period ($T$) is the time taken for one complete cycle of a periodic waveform to occur. It is the reciprocal of frequency ($T = 1/f$) and is typically measured in seconds (s).

(iii) Root-mean-square (RMS) value

• The Root-mean-square (RMS) value of a periodic waveform (like voltage or current) is the effective value that would produce the same average power dissipation in a resistive load as a DC voltage or current of the same magnitude. For a sinusoidal waveform, the RMS value is the peak value divided by $\sqrt{2}$.