Given the voltage and current shown in Fig. 16.70, find the parallel network internal to the container. That is, find the actual value of each component using the provided frequency. The image depicts a phasor diagram of an electrical component. It includes the following elements:1. **Current Expression:** - The current \( i \) flowing into the component is given by: \[ i = 6 \times 10^{-3} \sin(10,000t + 30^\circ) \] - This expression represents the instantaneous current as a sinusoidal function, where the amplitude is \( 6 \times 10^{-3} \) amperes, the angular frequency is \( 10,000 \) rad/s, and the phase angle is \( 30^\circ \).2. **Voltage Expression:** - The voltage \( v \) across the component is given by: \[ v = 24 \sin(10,000t - 40^\circ) \] - This expression represents the instantaneous voltage as a sinusoidal function, where the amplitude is \( 24 \) volts, the angular frequency is \( 10,000 \) rad/s, and the phase angle is \( -40^\circ \).3. **Phasor Diagram:** - The diagram includes a rectangular box labeled with a question mark, indicating an unknown component. - There are terminals on the left side of the box marked with “+” and “-”, representing the positive and negative sides of the component where the voltage \( v \) is measured. - An arrow labeled with current \( i \) enters the box from the left, indicating the direction of current flow into the component.This setup is typically used to analyze AC circuits, where the goal might be to determine the impedance of the unknown component by comparing the phase and magnitude of the current and voltage.

The relation between the voltage, current, and impedance using the ohm's law is as below Volage V = Current i * impedance Z here,Voltage v = 24sin(10000t - 40) = 242∠-40Current i = 6 * 10-3 sin(10000t - 40) = 6 * 10-3 2∠30impedance Z = Vi = 242∠-406* 10-32∠30 = 4000∠-70 Parallel network required R-C parallel circuit elements Admitttance Y = 1R+j1Xc---------(1)Also,Admittance Y = 1Z = 14000∠70 = 8.55 * 10-5 + j 2.35 * 10-4 S -----(2)Comparing the equation 1 and 21R = 8.55 * 10-5 ==> R = 11.695 kohm1Xc = 11wC= 2.35 * 10-4 ==> C = 23.5 nF