DC Electrical Circuits: Emf, Resistors, Kirchhoff's Rules, RC Circuits - Prof. Juyang Huan, Study notes of Physics

Download DC Electrical Circuits: Emf, Resistors, Kirchhoff's Rules, RC Circuits - Prof. Juyang Huan and more Study notes Physics in PDF only on Docsity! Chapter 28 Direct Current circuits 28.1 Electromotive Force (emf) Sources of emf: A battery or any other device that provides electrical energy. “charge pump” r: internal resistance Ir: potential decrease by r IR = V: potential on R, or terminal voltage of the battery V = - Ir, or = IR +Ir • is the open-circuit voltage. (if I = 0, V = = Vmax) • If r is small (a good battery), V . • If r is large (a bad battery), V << . • Power dissipation I = I 2R + I2r + _ Resistor Example: A battery has an emf of 12 V, and internal resistance of 0.05 . Its terminals are connected to a load resistance of 3 . (a)Find the current. I = R + r = 12V 3.05 = 3.93A (b) Find the terminal voltage: V = Ir = 12V (3.93A)(0.05 ) = 11.8V Find the equivalent resistance 6 8 4 3 12 2 14 28.2 Resistors in Series and in Parallel In Series: The current is the same through each resistor in series connection: I = I1 = I2 The total voltage = sum of voltages of each resistor: V =V1 +V2 = IR1 + IR2 = I(R1 + R2 ) = IReq Generally, Req = R1 + R2 + R3 + ... R1 I In Parallel: The voltage (potential difference) across each resistor in a parallel circuit is the same: V = V1 = V2 The total current = sum of currents passing through each resistor in a parallel circuit I = I1 + I2 = V R1 + V R2 = V 1 R1 + 1 R2 = v Req Generally, 1 Req = 1 R1 + 1 R2 + 1 R3 + ... Water in the bucket Charge in the capacitor Water flow in the pipe Electric current q t C 0.632 I t I0 0.368I0 I0 = R R S I +q -q t flow rate 0.368F0 Water in bucket t Wmax 0.632 Bucket pipe Reservoir 28.4 RC Circuits Charging a capacitor: t < 0: C is NOT charged and I = 0 t = 0: The switch is closed. t > 0: C begins to charge. When t > 0, at any given moment, Kirchhoff’s loop rule gives: IR q C = 0 (1) Initial Current: At t = 0, q = 0. No potential drop on C, all the potential drop is on R. Therefore, I0 = R (current at t = 0) (2) When t , C is charged to its maximum qmax, and the current is 0. All the potential drop is on C. No potential drop on R. qmax = C (maximum charge) (3) Between the above two limiting cases, the charge is governed by eq: dq dt = R q RC The solution: q(t) = C [1 e t / RC ] = qmax[1 e t / RC ] C R S t < 0 R S t > 0 I +q -q I(t) = R e t / RC = I0e t / RC Time constant: = RC, has a unit of seconds. At t = , I = e 1I0 = 0.368I0 , q = qmax[1 e 1] = 0.632qmax = 0.632C The work done by the battery during the charging: Wbattery = Qmax = C 2 When C is fully charge, the energy stored in C: U = U = 12 qmax = 1 2C 2 = 12Wbattery The other half of energy goes into joule heat in the resistor. q t C 0.632C I t I0 0.368I0 I0 = R = RC Discharging a Capacitor Kirchhoff’s loop rule: q/c –IR = 0 IR = q c (1) Initial Current: At t = 0, q = Q I0 = Q RC (current at t = 0) (2) When t , I 0, q 0 All the energy become heat. (3) Generally, dq dt = q RC Solve it: I(t) = I0e t / RC q(t) =Qe t / RC q I 0.368Q t Q t I0 0.368I0 I0 = Q RC C R S t < 0 +Q -Q C R S t > 0 +q -q I Example: Charging a capacitor, = 12 V, C = 5 μF, R = 8 105 . (a) Find time constant = RC = (8 105 )(5 10-6 F) = 4.0 s (b) Find the qmax Q = C = (5 10-6 F)( 12 V) = 60 10-6 C = 60 μC (c) Find the maximum current I0 = /R = 12 V/ 8 10 5 = 15 10-6 A = 15 μA (d) I(t) and q(t) q(t) = 60.0 [1 e t/4 ] μC I(t) = 15.0 e t/4 μA (e) Calculate q(t) and I(t) at t =1 time constant Let t = 1 = 4 s q(t) = 60.0 [1 e 1] μC = 60.0 0.632 μC = 37.9 μC I(t) = 15.0 e 1 μA = 15.0 0.368 μA = 5.52 μA (f) After how many time constants is the current one tenth of its initial value? Let I(t) = I0 10 = I0 e t/ 1 10 = e t/ -ln10 = t t = ln10 = 2.30 C R S