A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude V and angular frequency ω. Let the source voltage be given by v=Vcosωt.

A) Show that the instantaneous voltages vR,vL, and vC at any instant are each equal to v and that i=iR+iL+iC, where i is the current through the source and iR, iL, and iC are the currents through the resistor, the inductor, and the capacitor, respectively.

B)What is the phase of iR with respect to v?

C)What is the phase of iL with respect to v?

D)What is the phase of iC with respect to v?

E)Draw a phasor diagram for the currents i, iR, iL, iC.

F)Use the phasor diagram of part (e) to find the current amplitude I for the current i through the source. Express your answer in terms of the variables IR, IC and IL .

G)Show that the result of part (f) can be written as I=V/Z, with 1/Z=sqrt(1/R^2+(ωC−1/ωL)^2).

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A)When the resistor, inductor and the capacitor are connected parallel to the source ' \mathrm{v} ', the potential difference across each of them is equal to the potential difference across the source.Hence the potential difference across each of them will be equal to potential difference of source that is v=v \cos \omega tTherefore,V_{R}=V_{C}=V_{L}=V=v \cos \omega tAlso, since they are parallel, the current passing through the source will be divided and will pass through each branch.Therefore, according to Kirchhoff's Junction ruleI=I_{R}+I_{C}+I_{L}B)The phase difference of with respect to source voltage is-For resistor,I_{R}=\frac{V \cos \omega t}{R}therefore, \phi=0Phase angle \phi=0 C)For inductor,\begin{array}{l}V_{L}=V \cos \omega t=L \frac{d i}{d t} \\\frac{V}{L} \int \cos \omega t d t=\int d i \\\frac{V}{L \omega}(\sin \omega t)=I_{L} \\\frac{V}{X_{L}} \cos \left(\omega t-\frac{\pi}{2}\right)=I_{L}\end{array}Therefore, \phi=-\frac{\pi}{2} D)For capacitor,\begin{array}{l} V_{C}=V \cos \omega t=\frac{Q}{C} \\I_{C}=\frac{d Q}{d t} \\I_{C}=\frac{C V \cos \omega t}{d t} \\I_{C}=C V \omega(-\sin \omega t) \\I_{C}=\frac{V}{\frac{1}{\omega C}} \cos \left(\omega t+\frac{\pi}{2}\right) \\I_{C}=\frac{V}{X_{C}} \cos \left(\omega t+\frac{\pi}{2}\right) \\\text { for } I_{C}, \phi=\frac{\pi}{2}\end{array} E)Phasor diagram for parallel RLC circuitF) From the phasor diagramI^{2}=I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}} G)Using Equation (1)\begin{array}{l}I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}} \\I=\sqrt{\frac{V^{2}}{R^{2}}+\left(\frac{V}{X_{C}}-\frac{V}{X_{L}}\right)^{2}} \\I=\frac{V}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}} \\I=\frac{V}{Z}\end{array}WhereZ=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}} ...