4 resistors whose resistances are 2k, 3k, 6k in Parallel. Determine power supply of AC 240V. Determine Total resistance. Determine Total Current. Determine Power developed in the circuit.

The formula for parallel resistors is:

$$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$$

Substitute the given resistance values:

$$\frac{1}{R_T} = \frac{1}{2 \, \text{k}\Omega} + \frac{1}{3 \, \text{k}\Omega} + \frac{1}{6 \, \text{k}\Omega} + \frac{1}{8 \, \text{k}\Omega}$$

Find a common denominator, which is 24:

$$\frac{1}{R_T} = \frac{12}{24 \, \text{k}\Omega} + \frac{8}{24 \, \text{k}\Omega} + \frac{4}{24 \, \text{k}\Omega} + \frac{3}{24 \, \text{k}\Omega}$$ $$\frac{1}{R_T} = \frac{12 + 8 + 4 + 3}{24 \, \text{k}\Omega} = \frac{27}{24 \, \text{k}\Omega}$$

$$R_T = \frac{24}{27} \, \text{k}\Omega = \frac{8}{9} \, \text{k}\Omega$$

Convert $R_T$ to Ohms for use with voltage in Volts:

$$R_T = \frac{8}{9} \times 1000 \, \Omega \approx 888.89 \, \Omega$$

The total resistance is $\boxed{888.89 \, \Omega}$.

$$I = \frac{V}{R_T}$$

Substitute the voltage and the total resistance (using the more precise fractional value for $R_T$):

$$I = \frac{240 \, \text{V}}{\frac{8000}{9} \, \Omega} = \frac{240 \times 9}{8000} \, \text{A}$$ $$I = \frac{2160}{8000} \, \text{A} = \frac{27}{100} \, \text{A} = 0.27 \, \text{A}$$

Using the rounded value $R_T \approx 888.89 \, \Omega$:

$$I = \frac{240 \, \text{V}}{888.89 \, \Omega} \approx 0.26999 \, \text{A}$$

Your calculation of $269.966 \times 10^{-3} \, \text{A}$ is approximately $0.270 \, \text{A}$.
The total current is $\boxed{0.270 \, \text{A}}$.

$$P = I \times V$$

Using the current value $I = 0.269966 \, \text{A}$ (from your calculation) and the given voltage $V = 240 \, \text{V}$:

$$P = (0.269966 \, \text{A}) \times (240 \, \text{V})$$ $$P = 64.79184 \, \text{W}$$

Rounding to two decimal places:

$$P = 64.79 \, \text{W}$$

The power developed in the circuit is $\boxed{64.79 \, \text{W}}$.