Exercise:
Consider the charging process in an RC circuit. abcliste abc Qualitatively describe the power delivered by the voltage supply the power delivered to the capacitor and the power dissipated in the resistor as a function of time. abc Derive formal expressions for the three powers in a and graph them. abc When does the power delivered to the capacitor reach its peak value? Calculate the peak power. abc Calculate the energy delivered by the voltage supply the energy stored in the capacitor and the energy dissipated in the resistor during the charging process. abcliste

Solution:
abcliste abc Since the power corresponds to the product of voltage and current we have to consider the behaviour of these to quantities for each of the three circuit elements. The current through the entire circuit is the same and exponentially decreases from the initial current to . The voltage supply delivers a constant voltage. Therefore the power delivered by the voltage supply exponentially decreases with the same time constant as the current. The voltage across the resistor exponentially decreases with the same time constant as the current. Therefore the power also decreases exponentially but with a shorter time constant. This can be seen since after one half-life of the current decrease the power has dropped to one quarter of its inital value. The voltage across the capacitor starts from and asymptotically ts towards the voltage provided by the voltage supply. The product of voltage and current starts from and ts to for large times. We can therefore expect it to have a peak between the ning and the of the charging process. abc The current is given by It I_ e^-t/tau The power deliverd by the voltage supply is then sscPtott Delta V_ It Delta V_ I_ e^-t/tau PtF with the initial power P_Delta V_ I_. medskip For the resistor we find P_Rt Delta V_R It Delta V_ e^-t/tau I_ e^-t/tau P_lefte^-t/tauright^ PRF with the time constant tau'tau/. This means that the power decreses twice as fast as the current i.e. with half the time constant. medskip The voltage across the capacitor is given by Delta V_Ct Delta V_ left- e^-t/tauright It follows for the power P_Ct Delta V_ left- e^-t/tauright I_ e^-t/tau P_lefte^-t/tau-lefte^-t/tauright^right labelpcap PCF It is obvious that the power delivered by the voltage supply is equal to the of the power dissipated in the resistor and the power delivered to the capacitor: sscPtott P_Rt+P_Ct labelptot A typical example of the power vs. time graphs can be seen below. center includegraphicswidthcm#image_path:rc-circuit-power# center abc In order to find the peak power we can derive the power for the capacitor with respect to time: dvP_Ctt dvleftPCFrightt P_left-fractau e^-t/tau+fractau' e^-t/tau'right At the maximum the derivative has to be zero. It follows that fractau'tau frace^-t/tau'e^-ttau fraclefte^-t/tauright^e^-t/taue^-t/tau Since fractau'tau frac this is exactly the condition for the half-life i.e. the peak power is reached after tT_/tau ln . Plugging ttau ln o refpcap leads to sscP_Cmax P_Ctauln P_lefte^-tauln /tau-lefte^-tauln /tauright^right P_lefte^-ln -lefte^-ln right^right P_leftfrac-leftfracright^right PmaxF abc The instantaneous power can be written as the derivative of energy work with respect to time: Pt dvWt This also means that the energy corresponds to the egral of power with respect to time: W _t_^t_Ptddt For the total energy delivered by the voltage supply we find sscWtot _^inftysscPtottddt _^inftyPtFddt left-P_tau e^-t/tauright_^infty -P_taulefte^-t/tauright_^infty -P_tauleft-right WtF For the energy dissipated in the resistor we get W_R _^inftyP_Rtddt _^inftyPRFddt This is the same egral as before just with a different time constant. Therefore the energy is W_R P_tau' fracsscWtot Because of refptot it is clear that the remaining energy is delivered to the capacitor: W_C W_R fracsscWtot labelwcap fracWtF fracDelta V_ I_ R C fracDelta V_^ R C R fracC Delta V_^ This is the well-known result for the energy stored in a capacitor. medskip It is a good exercise to derive refwcap by egrating refpcap. medskip In other words when charging a capacitor using an RC circuit only half of the energy provided by the voltage supply is stored in the capacitor while the other half is disspiated in the resistor. The result does not dep on any specifics of the circuit. There are fortunately more efficient ways to charge a capacitor. abcliste