General Physics (PHY 2140) Lecture 6 Electrodynamics Direct current circuits parallel and series connections Kirchhoff’s rules RC circuits http://www.physics.wayne.edu/~alan/2140Website/Main.htm Chapter 18 9/21/2018

The Physics Resource Center Room 172 of Physics Research Building. Department of Physics and Astronomy announces the Spring-Summer 2007 opening of The Physics Resource Center on Monday, May 21 in Room 172 of Physics Research Building. Hours of operation: Monday and Tuesday 10:00 AM to 5:00 PM Wednesday and Thursday 10:00 AM to 4:00 PM Friday, Saturday and Sunday Closed Undergraduate students taking PHY1020, PHY2130, PHY2140, PHY2170/2175 and PHY2180/2185 will be able to get assistance in this Center with their homework, labwork and other issues related to their physics course. The Center will be open: Monday, May 21 to Thursday, August 2, 2007. 9/21/2018

Lightning Review Last lecture: Current and resistance Temperature dependence of resistance Power in electric circuits DC Circuits EMF Resistors in series 9/21/2018

Introduction: elements of electrical circuits A branch: A branch is a single electrical element or device (resistor, etc.). A junction: A junction (or node) is a connection point between two or more branches. 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).      A circuit with 5 branches.    A circuit with 3 nodes. 9/21/2018

18.1 Sources of EMF 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” I V = IR E + - V = IR = E 9/21/2018

EMF E 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 DV is Note: E is the same as the terminal voltage when the current is zero (open circuit) B C r E R A D 9/21/2018

EMF (continued) E 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 Thus, the current in the circuit is B C r R E A D Power output: 9/21/2018 Note: we’ll assume r negligible unless otherwise is stated

Measurements in electrical circuits 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 9/21/2018

Direct Current Circuits Two Basic Principles: Conservation of Charge Conservation of Energy Resistance Networks Req I a b Ohm’s Law: 9/21/2018

18.2 Resistors in series 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, A B I 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, C By definition, Thus, Req would be 9/21/2018

Resistors in series: notes Analogous formula is true for any number of resistors, It follows that the equivalent resistance of a series combination of resistors is greater than any of the individual resistors (series combination) 9/21/2018

Resistors in series: example 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 B with I Then, C Thus, This circuit is known as voltage divider. 9/21/2018

18.3 Resistors in parallel 1. Since both R1 and R2 are connected to the same battery, potential differences across R1 and R2 are the same, A 2. Because of the charge conservation, current, entering the junction A, must equal the current leaving this junction, By definition, Thus, Req would be or 9/21/2018

Resistors in parallel: notes Analogous formula is true for any number of resistors, It follows that the equivalent resistance of a parallel combination of resistors is always less than any of the individual resistors (parallel combination) 9/21/2018

Resistors in parallel: example 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: with Then, Thus, This circuit is known as current divider. 9/21/2018

Direct current circuits: example Find the currents I1 and I2 and the voltage Vx in the circuit shown below. Strategy: Find current I by finding the equivalent resistance of the circuit Use current divider rule to find the currents I1 and I2 Knowing I2, find Vx. 9/21/2018

Direct current circuits: example Find the currents I1 and I2 and the voltage Vx in the circuit shown below. First find the equivalent resistance seen by the 20 V source: Then find current I by, We now find I1 and I2 directly from the current division rule: Finally, voltage Vx is 9/21/2018

R1=4W E=18V R3=6W R2=3W Example: Determine the equivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery R1=4W + E=18V R2=3W R3=6W 9/21/2018

R1=4W R1=4W E=18V R23=2W R3=6W R2=3W So, Ieq= E/Req=3A E=18V Req=6W Example: Determine the equivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery R1=4W R1=4W + + E=18V R23=2W R2=3W R3=6W + So, Ieq= E/Req=3A P=IeqE = 108W E=18V Req=6W 9/21/2018

18.4 Kirchhoff’s rules and DC currents 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). 9/21/2018

1. Junction rule 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. • Similar to the water flow in a pipe. 9/21/2018

2. Loop rule • Assign symbols and directions of currents in the loop As a consequence of the Law of the conservation of energy, we have: The sum of the potential differences across all the elements around any closed loop must be zero. • Assign symbols and directions of currents in the loop If the direction is chosen wrong, the current will come out with the correct 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 9/21/2018

Simplest Loop Rule Example: Single loop, start at point A Battery traversed from – to + So use + E (a voltage gain) Resistor traversed from + to – So use – IR (a voltage drop) For the loop we have: I B C + + E R - - A D 9/21/2018

Loop rule: more complex illustration 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. + - v2 - v5 + - - - Path 1 v1 v4 v6 + + + Path 2 v3 v7 - + + - Path 3 - + + v8 v12 v10 + - - 9/21/2018 + v11 - - v9 +

Loop rule: illustration “b” Using sum of the drops = 0 • + - v2 - v5 + - - - Blue path, starting at “a” + v7 - v10 + v9 - v8 = 0 v1 v4 v6 + + + v3 v7 - + + - • “a” Red path, starting at “b” -v2 + v5 + v6 + v8 - v9 + v11 + v12 - v1 = 0 - + + v8 v12 v10 + - - Yellow path, starting at “b” - v2 + v5 + v6 + v7 - v10 + v11 + v12 - v1 = 0 + v11 - - v9 + 9/21/2018

Kirchhoff’s Rules: Single-loop circuits Example: For the circuit below find I, V1, V2, V3, V4 and the power supplied by the 10 volt source. For convenience, we start at point “a” and sum voltage drops =0 in the direction of the current I. + _ V 1 4 3 2 30 V 10 V 15 W 40 5 20 20 V I "a" · -10 - V1 + 30 - V3 - V4 + 20 - V2 = 0 (1) 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 (3) 9/21/2018 Solving this equation gives, I = 0.5 A

Kirchhoff’s Rules: Single-loop circuits (cont.) + - _ V 1 4 3 2 30 V 10 V 15 W 40 5 20 20 V I "a" · Using this value of I in Eq. 2 gives: V1 = 10 V V3 = 7.5 V V2 = 20 V V4 = 2.5 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. 9/21/2018

18.5 RC circuits When switch is closed, current flows because capacitor is charging As capacitor becomes charged, the current slows because the voltage across the resistor is E - Vc and Vc gradually approaches E. Once capacitor is charged the current is zero Q 0.63 Q Charge across capacitor RC is called the time constant 9/21/2018

Discharging the capacitor in RC circuit If a capacitor is charged and the switch is closed, then current flows and the voltage on the capacitor gradually decreases. This leads to decreasing charge Q 0.37Q Charge across capacitor 9/21/2018

Example : charging an unknown capacitor A series combination of a 12 kW resistor and an unknown capacitor is connected to a 12 V battery. One second after the circuit is completed, the voltage across the capacitor is 10 V. Determine the capacitance of the capacitor. 9/21/2018

A series combination of a 12 kW resistor and an unknown capacitor is connected to a 12 V battery. One second after the circuit is completed, the voltage across the capacitor is 10 V. Determine the capacitance of the capacitor. I R C E Given: R =12 kW E = 12 V V =10 V t = 1 sec Find: C=? Recall that the charge is building up according to Thus the voltage across the capacitor changes as This is also true for voltage at t = 1s after the switch is closed, 9/21/2018

Magnetic field pattern for a horse-shoe magnet Chapter 19: Magnetism Magnetic field pattern for a horse-shoe magnet 9/21/2018