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General Physics (PHY 2140) Lecture 10 Electrodynamics Direct current circuits parallel and series connections Kirchhoff’s rules Chapter 18 5/23/2017 1 Last lecture: 1. Current and resistance Q I t Temperature dependence of resistance Power in electric circuits R Ro 1 T To V P I V I 2 R 2 R Review Problem: Consider a moose standing under the tree during the lightning storm. Is he ever in danger? What could happen if lightning hits the tree under which he is standing? 5/23/2017 2 A branch: A branch is a single electrical element or device (resistor, etc.). A circuit with 5 branches. A junction: A junction (or node) is a connection point between two or more branches. A circuit with 3 nodes. If we start at any point in a circuit (node), proceed through connected electric devices back to the point (node) from which we started, without crossing a node more than one time, we form a closed-path (or loop). 5/23/2017 3 Steady current (constant in magnitude and direction) • requires a complete circuit • path cannot be only resistance cannot be only potential drops in direction of current flow Electromotive Force (EMF) • provides increase in potential E • converts some external form of energy into electrical energy Single emf and a single resistor: emf can be thought of as a “charge pump” V = IR I + - V = IR = E E 5/23/2017 4 Each real battery has some internal resistance AB: potential increases by E on the source of EMF, then decreases by Ir (because of the internal resistance) Thus, terminal voltage on the battery V is V E Ir B C r R E A D Note: E is the same as the terminal voltage when the current is zero (open circuit) 5/23/2017 5 Now add a load resistance R Since it is connected by a conducting wire to the battery → terminal voltage is the same as the potential difference across the load resistance V E Ir IR, or E Ir IR Thus, the current in the circuit is I 5/23/2017 E Rr B C r R E A D Power output: I E I 2r I 2 R Note: we’ll assume r negligible unless otherwise is stated 6 Voltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device. V Ammeters measure current through a device by being placed in series with the device. A 5/23/2017 7 Two Basic Principles: Conservation of Charge Conservation of Energy a Resistance Networks Vab IReq Req 5/23/2017 Vab I I Req b 8 A + v2 _ B R2 + v _ + Ii1 R1 1. Because of the charge conservation, all charges going through the resistor R2 will also go through resistor R1. Thus, currents in R1 and R2 are the same, I1 I 2 I v1 _ C 2. Because of the energy conservation, total potential drop (between A and C) equals to the sum of potential drops between A and B and B and C, V IR1 IR2 By definition, V IReq Thus, Req would be Req V IR1 IR2 R1 R2 I I Req R1 R2 5/23/2017 9 Analogous formula is true for any number of resistors, Req R1 R2 R3 ... (series combination) It follows that the equivalent resistance of a series combination of resistors is greater than any of the individual resistors 5/23/2017 10 In the electrical circuit below, find voltage across the resistor R1 in terms of the resistances R1, R2 and potential difference between the battery’s terminals V. Energy conservation implies: A + v2 _ with V V1 V2 V1 IR1 and V2 IR2 v1 Then, V I R1 R2 , so I C Thus, B R2 + v _ + Ii1 R1 _ V1 V V R1 R2 R1 R1 R2 This circuit is known as voltage divider. 5/23/2017 11 I A I2 + I I1 V R2 _ 1. Since both R1 and R2 are connected to the same battery, potential differences across R1 and R2 are the same, R1 + V V1 V2 V Req 2. Because of the charge conservation, current, entering the junction A, must equal the current leaving this junction, _ I I1 I 2 V Req By definition, I Thus, Req would be V1 V2 V V V I Req R1 R2 R1 R2 5/23/2017 1 1 1 Req R1 R2 or Req R1 R2 R1 R2 12 Analogous formula is true for any number of resistors, 1 1 1 1 ... (parallel combination) Req R1 R2 R3 It follows that the equivalent resistance of a parallel combination of resistors is always less than any of the individual resistors 5/23/2017 13 In the electrical circuit below, find current through the resistor R1 in terms of the resistances R1, R2 and total current I induced by the battery. Charge conservation implies: I I1 I 2 I + I2 I1 with V _ R2 R1 Then, Thus, V V , and I 2 R1 R2 IReq RR I1 , with Req 1 2 R1 R1 R2 I1 I1 I R2 R1 R2 This circuit is known as current divider. 5/23/2017 14 Find the currents I1 and I2 and the voltage Vx in the circuit shown below. I Strategy: 7 + 20 V + _ Vx _ 5/23/2017 I2 I1 4 12 1. Find current I by finding the equivalent resistance of the circuit 2. Use current divider rule to find the currents I1 and I2 3. Knowing I2, find Vx. 15 Find the currents I1 and I2 and the voltage Vx in the circuit shown below. I 7 + 20 V + _ Vx I2 I1 4 12 _ Then find current I by, First find the equivalent resistance seen by the 20 V source: 4(12) Req 7 10 12 4 20V 20V I 2A Req 10 We now find I1 and I2 directly from the current division rule: I1 2 A(4) 0.5 A, and I 2 I I1 1.5 A 12 4 Finally, voltage Vx is Vx I 2 4 1.5 A 4 6V 5/23/2017 16 • • 1. 2. The procedure for analyzing complex circuits is based on the principles of conservation of charge and energy They are formulated in terms of two Kirchhoff’s rules: The sum of currents entering any junction must equal the sum of the currents leaving that junction (current or junction rule) . The sum of the potential differences across all the elements around any closed-circuit loop must be zero (voltage or loop rule). 5/23/2017 17 As a consequence of the Law of the conservation of charge, we have: • The sum of the currents entering a node (junction point) equal to the sum of the currents leaving. Ia Id Ic Ib I a + Ib = Ic + Id Similar to the water flow in a pipe. I a, I b, I c , and I d can each be either a positive or negative number. 115/23/2017 18 As a consequence of the Law of the conservation of energy, we have: • 1. Assign symbols and directions of currents in the loop – 2. The sum of the potential differences across all the elements around any closed loop must be zero. If the direction is chosen wrong, the current will come out with a right magnitude, but a negative sign (it’s ok). Choose a direction (cw or ccw) for going around the loop. Record drops and rises of voltage according to this: – – – – If a resistor is traversed in the direction of the current: +V = +IR If a resistor is traversed in the direction opposite to the current: -V=-IR If EMF is traversed “from – to + ”: +E If EMF is traversed “from + to – ”: -E 115/23/2017 19 Loops can be chosen arbitrarily. For example, the circuit below contains a number of closed paths. Three have been selected for discussion. Suppose that for each element, respective current flows from + to - signs. - + v 2 - v5 + - v1 v4 + v6 + - v3 Path 1 + Path 2 + v7 - + Path 3 v12 v10 - - 5/23/2017 v8 + + + + v11 - - v9 + 20 “b” • - Using sum of the drops = 0 + v 2 - v1 v4 + v3 v12 v10 - + v11 - - v7 + v10 – v9 + v8 = 0 • “a” v8 + + + Blue path, starting at “a” + + v7 - + v6 + - 5/23/2017 - v5 + - v9 + Red path, starting at “b” +v2 – v5 – v6 – v8 + v9 – v11 – v12 + v1 = 0 Yellow path, starting at “b” + v2 – v5 – v6 – v7 + v10 – v11 - v12 + v1 = 0 21 Example: For the circuit below find I, V1, V2, V3, V4 and the power supplied by the 10 volt source. 30 V + + _ V1 _ 10 V _ 20 1. "a" + _ _ 15 V3 40 I + V2 + 5 _ _ + V4 For convenience, we start at point “a” and sum voltage drops =0 in the direction of the current I. +10 – V1 – 30 – V3 + V4 – 20 + V2 = 0 (1) + 20 V 2. We note that: V1 = - 20I, V2 = 40I, V3 = - 15I, V4 = 5I (2) 3. We substitute the above into Eq. 1 to obtain Eq. 3 below. 10 + 20I – 30 + 15I + 5I – 20 + 40I = 0 5/23/2017 Solving this equation gives, I = 0.5 A. (3) 22 30 V + + _ V1 _ 10 V _ 20 + _ V3 Using this value of I in Eq. 2 gives: "a" _ 15 40 I + V2 V1 = - 10 V V3 = - 7.5 V V2 = 20 V V4 = 2.5 V + 5 _ + V4 _ + 20 V P10(supplied) = -10I = - 5 W (We use the minus sign in –10I because the current is entering the + terminal) In this case, power is being absorbed by the 10 volt supply. 5/23/2017 23 Consider the circuit switch E i t q 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.340 0.36 0.38 0.4 q C +q -q vc = q/C R vR = iR q Q 1 e 5/23/2017 t RC 0.2 i 0 1 0.064493015 0.935507 0.124826681 0.8751733 0.181269247 0.8187308 0.234071662 0.7659283 0.283468689 0.7165313 0.329679954 0.67032 0.372910915 0.6270891 0.41335378 0.5866462 0.451188364 0.5488116 0.486582881 0.5134171 0.519694699 0.4803053 0.550671036 0.449329 0.579649615 0.4203504 0.606759279 0.3932407 0.632120559 0.3678794 0.655846213 0.3441538 0.678041728 0.4 0.60.3219583 0.8 0.698805788 0.3011942 0.718230711 0.2817693 0.736402862 0.2635971 t i 1 1.2 RC is called the time constant 24 Discharge the capacitor t switch E i q 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.340 0.36 0.38 0.4 q C +q -q vc = q/C R vR = iR 0.2 i 1 1 0.935506985 0.935507 0.875173319 0.8751733 0.818730753 0.8187308 0.765928338 0.7659283 0.716531311 0.7165313 0.670320046 0.67032 0.627089085 0.6270891 0.58664622 0.5866462 0.548811636 0.5488116 0.513417119 0.5134171 0.480305301 0.4803053 0.449328964 0.449329 0.420350385 0.4203504 0.393240721 0.3932407 0.367879441 0.3678794 0.344153787 0.3441538 0.321958272 0.4 0.60.3219583 0.8 0.301194212 0.3011942 0.281769289 0.2817693 0.263597138 0.2635971 t i 1 1.2 q Qet RC 5/23/2017 25