We continue from the rectifier circuit in Fig. $1.$ Given that V p $=2302$ V and the line angular frequency $0=2$$50.$ The load resistor has the value R $=50.$ Fig. $5$ shows the output voltage waveform of the rectifier. The pulsating output of the diode rectification $($green$)$ is smoothen by a filtering capacitor at the output $($red$).$ The result is a DC voltage signal with a voltage ripple $$v $=16\mathrm{.}26$ V$.$ It is assumed that the capacitor value is large enough such that starting at $=$ $2$ the output voltage is supported by the discharging capacitor instead of the source AC voltage. The diodes are considered ideal diodes $($no forward voltage drop, no extra resistance, $...).$ Figure $5$: Output waveform of the diode rectifier. $($a$)$ Write the output voltage expression as a function of the phase angle $,$ using the given parameters. Use only the parameter names instead of substituting their actual values into the equations. $($Hint: C is discharging after $/2.)($b$)$ Find the value of $1.($c$)$ Find the value of the filtering capacitor such that the given voltage ripple condition is satisfied. $($d$)$ Calculate the load resistor average power consumption $($per source voltage period$)$ using the average power equation P $=1$ T Z T $0$ v$($t$)$i$($t$)$ dt $=12$ Z $2$ $0$ v$($$)$i$($$)$ d $2/5$ TAU EE$.$PEE$\mathrm{.}210$ Exercise $2($e$)$ Calculate the average DC voltage level using the average signal equation V $=1$ T Z T $0$ v$($t$)$ dt $=12$ Z $2$ $0$ v$($$)$ d $-$