Question

Consider the RLC circuit shown in the figure. R L V Select 'True', 'false' or 'Cannot tell' for the following statements. The voltage drop across the resistor is the same as the voltage drop across the inductor at all times. OA: Truel OB: False OC: Cannot tell The current through the inductor always equals the current charging/discharging the capacitor. OA: True OB: False OC: Cannot tell The current through the inductor is the same as the current through the resistor at all times. OA: Truel OB: False OC: Cannot tell Energy is dissipated in the resistor but not in either the capacitor or the inductor. OA: Truel OB: False OC: Cannot tell

Answer 1

С. R son If w Let, i is the instantenious current through the circuit . And no is the Voltage of the ac source. Ldi I С Now, potential drop across the inductance = VL dt Potential drop aeross the resistance, ez ir ve and the same aeross the capaeitor is 9 Sidt - Ne so from KVL I can while Side © cit Vol Krow, in rotating vector method we assumne (Here, 5= Fi) and Rit L di - + 10 C ایمان V = 9i= Toegwt +$) Therefore, di لاژ Le jwi and į lidt - zwi Therefore, Sidt = di TE świ 1 jw so, from equation - 9 can write Rit sulit ܂ ܢ ó WC i (R+ IWL + skva) = v ) لار N (3 .کے = R+i (EL-MG) T R+ JWL+ V jwc VR NRP This, is instantaneous current through lo ekt.circuit. voltage drop across the resistono Ri voltage dro aeross the induetor v =ówli Rtö(wL-tic) 10 با R+j(wL-Woc You can see voltage across resistor and voltage across the inductor is not same at all times. [False

We all of the three are in boies connection. So, Current thingh NOW the three components will always be same. 'R = iç-iz Coolways) Truel and Tone Energy dissipration : - You know average power defined as P= 1 T 1 / WH) i(t) dt now let the phase angle between current and voltage is $ and also assume v (t) = No sinut and ilt) = I Sin (wt+o) P- — v. to join ut. Sin fot + d) de wat seat [eoso - cos(zut + d)Jdt No to east 2 Va re To va cost Vph case Hese Ny and Iy are the re.m.s. value of vellage and ewerent. Now from ean - you can see the impedance ZER+ j(WL-60) of a tan' (WL and 121= J R² + (2W - 1 wc.)2 = Izleit | wo 50 v R (4) R so, coso VR2+ (wLýc)? Here & is the phase angle by which current lags behind the voltage So, Power in the circuit P= Vy Ip lost eqh - (3) Vp = In VR²+ (W-1cd2 Now if you use eqh (4) and eq"-6 then you can have P= IR Again from

This indicates that Energy is dissipated in the resistor only. You can see that the current will be maximum when the denominator is True from ean ③ you can write lil= | | WC Now, JR?+ (WL-1 wc)² minimum. ie. WL WO - La W²C (2TfJ2C H (2021.71X163 *6.1*162 = 1.42x16GH This is called Resonance condition. In this condition the impedance 121= R2+(WL-buc) R²+0² - R > and also Phorse angle o=0 so Z=R Impedance seen by the voltage source is the resistance R 2.25 X 10²2 R = 2.25