4 PHY167 Spring Electric Circuits 2021

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4 – Electric circuitsPHY167 Spring 2021Serial and parallel resistorsSerial connection of resistors:V1V2V3R1R2R3As the current I through each of serially connected resistors is the same, one can useOhm’s law and writeV V1 V2 V3 . R1I R2 I R3 I . ( R1 R2 R3 ) IThat is, one can consider serially connected resistors as one combined resistance:V RIThusR R1 R2 R3 .1

DVParallel connection of resistors:R1R2R3Here the potential difference V is common for all resistors, whereas total current is the sum ofindividual currents:𝑉𝑉𝑉𝐼 𝐼1 𝐼2 𝐼3 𝑅1 𝑅2 𝑅3The total resistance is defined asI VRThus1 111 .R R1 R2 R31/R can be called conductance. In the case of parallel connection, conductances add up.2

VSerial and parallel capacitorsC1C2Parallel connection of capacitors:C3Several capacitors connected in parallel form an effective capacitor C whose charge isthe sum of the charges on all capacitors, whereas the voltage is common:Q Q1 Q2 Q3 . C1V C2V C3V C1 C2 C3 V CVThus for the parallel connection of capacitors one obtainsC C1 C2 C3Serial connection of capacitors:VV1V2V3Here voltages add up whereas the charge is common (on each of the capacitors the same)V V1 V2 V2 . 1Q Q Q11 Q . Q C1 C2 C3 C1 C2 C3 CThus for the serial connection of capacitors one obtains1 111 .C C1 C2 C33

Kirchhoff’s laws1. At any junction point, the sum of all currents entering the junction must be equal thesum of all currents leaving the junctionI1I3I2I4I1 I2 I3 I4(Electric charges are conserved andare not accumulating in the wire)2. For any closed path in the circuit, the change of its electric potential around the path iszero.V1 V 2 V3 V 4 0V1V4(Electric field is conservative)V2V3An equivalent formulation of the 2nd Kirchhoff’s law:For any path between the two points in a circuit, thesum of all voltages on the path is equal to thevoltage V between these two points:𝑉𝑖 𝑉𝑖4

ProblemR1R4Calculate the effective resistanceof this circuit. Obtain its numericalvalue forR1 1 W, R2a 2 W, R2b 1 W, R3 3 W, R4 1 W, R5 1 W,R2aR2bR5R3Solution: We at first replace the serially connected resistances R2aand R2b by the effectiveresistance R2 R2a R2b. Then we replace the central group of parallel connected resistances by the effectiveresistance 111 R1 R2 R3 1The last step is to replace the three serially connected resistances by the final effectiveresistance: 1 111 R R4 R5RR RR3 2a2b 1Plugging the numbers: 111 1R 1 1 2.6 W 1 2 1 3 5

The Wheatstone bridgeNot all circuits can be calculated using theformulas for the serial and parallel connection ofresistors. The simplest example is the so-calledWheatstone bridge. The Kirchhoff’s equations forthis circuit are the following.1st Kirchhoff:𝐼 𝐼1 𝐼3𝐼1 𝐼2 𝐼5𝐼3 𝐼5 𝐼42nd Kirchhoff Ohm:𝑅1 𝐼1 𝑅2 𝐼2 𝑉𝑅3 𝐼3 𝑅4 𝐼4 𝑉𝑅1 𝐼1 𝑅5 𝐼5 𝑅4 𝐼4 𝑉This is a system of 6 linear equations for 6 unknowns – all currents. It can be solved by computeralgebra. After finding all currents, one finds the effective resistance:𝑉𝑅1 𝑅2 𝑅3 𝑅4 𝑅5 𝑅1 𝑅3 𝑅2 𝑅4 𝑅2 𝑅4 𝑅1 𝑅3𝑅 𝐼𝑅1 𝑅2 𝑅3 𝑅4 𝑅5 𝑅1 𝑅3 𝑅2 𝑅4Limiting cases can be calculated easily and used to check the general formula above𝑅 𝑅1 𝑅3𝑅2 𝑅4 𝑅1 𝑅3 𝑅2 𝑅4𝑅5 0𝑅1 𝑅2 𝑅3 𝑅4𝑅 𝑅1 𝑅2 𝑅3 𝑅4𝑅5 6

Electromotive forceConsideration of closed electric circuits that consist of resistances only, like this oneV 0(According to the 2nd Kirchhoff’s law)shows that the electric current is zero. Indeed, there is no reason for the current to flowalong the closed loops in the circuit because the total change of the potential across anyloop is zero. What causes electric charges to flow are non-electric forces such as chemicalforces in batteries. These forces are not potential forces because the work done by theseforces along closed loops is nonzero. This is exactly the reason for the currents to flow inclosed electric circuits.Non-electric forces usually act in finite regions. In the case of batteries, they act only withinbatteries. The work done by non-electric forces on a test charge q that crosses the region ofaction of these forces, divided by q, is called Electromotive force and denoted by E. Notethat the term Electromotive force is misleading. First, it is of non-electric origin, although itmoves electric charges. Second, it is not force but a quantity resembling the electricpotential.7

Batteries, EMF F/q dE E Fd/qV EdFor an isolated (non-connected) battery the difference of the electric potential between theelectrodes (the voltage) is equal to the electromotive force, V E. That is, electric and non-electricforces acting on the charges compensate each other everywhere inside the battery, so that the netforce acting on a charge is zero and there is no electric current. The voltage on the battery arisesbecause the chemical forces F move electrons to the right (thus positive charges to the left) so thatelectrodes become charged positively and negatively, respectively.Change of the potential in a circuit with a battery. Potential increases across the battery anddrops on the resistors, if we are moving in the shown direction of the current. Charges(positive) are moving outside the battery down the potential, like skiers in the mountains. Insidethe battery they are moving up the potential under the influence of the electromotive force (analso electric force). The electromotive force plays the role of the ski lift.VVV R1 R3EVR28

Ohm’s law with EMFConsider a battery and a resistor connected serially. The resistor can stand for the internalresistance of the battery.𝐼 - the chosen positive direction of the current𝑉2 𝑉1Voltage on the system𝑉 𝑉1 𝑉2 𝑉upstream 𝑉downstreamEOhm’s law with EMF:𝑉 ℰ 𝑅𝐼𝐼 Consider the simplest closed circuit: battery and the load.Voltage on the load resistor R isVoltage on the battery is𝑉1 𝑉2 𝑉𝑉2 𝑉1 𝑉R0𝑉1RThus the Ohm’s laws readV RI V E R0 IMinus in front of V means that the voltage would movethe current across the battery in the negative direction,in the absence of the electromotive force E.Adding these equations yieldsE ( R R0 ) I𝑉2I ER R0V RI RER R09

2nd Kirchhoff’s law with EMFThe second Kirchhoff’s law states that for each closed loop in the circuit the sum of voltages iszero that reflects the fact that electric potential is defined unambiguously (and the work of theelectric field over each closed trajectory is zero):𝑉𝑖 0.𝑖To the Kirchhoff’s laws, one has to add the Ohm’s law𝑉𝑖 𝑅𝑖 𝐼𝑖for each resistor. On the top of it, there can be EMF’s acting within resistors (batteries have theirown internal resistance and thus can be considered as resistors) and pushing the current throughthem. With the EMF’s, the Ohms law becomes𝑉𝑖 E𝑖 𝑅𝑖 𝐼𝑖 .Substituting 𝑉𝑖 𝑅𝑖 𝐼𝑖 E𝑖 into the second Kirchhoff’s law, one obtainsE𝑖 .𝑅𝑖 𝐼𝑖 𝑖for each closed loop in a circuit.𝑖10

ProblemR0A 1.5 V battery with the internal resistance 5 W isconnected to a light bulb with a resistance of 20 W ina simple single-loop circuit. (a) What is the currentflowing in the circuit? (b) What is the voltagedifference across the light bulb?RLight bulbSolution: Use the results of the previous slideI E1.5 0.06 AR R0 5 20V IR 0.06 20 1.2 V11

ProblemA 1.5 V battery with the internal resistance 5 W isconnected to a light bulb with a resistance of 20 W ina simple single-loop circuit. (a) What is the powerdissipated in the circuit? (b) What is the powerdissipated in the light bulb and in the battery?R0RLight bulbSolution: (a) The total dissipated power in the circuit is equal to the power of theelectromotive force:E21.52P IE 0.09 WR R0 20 5(b) The power dissipated on the bulb isPbulbE2R1.52 20 I R 0.072 W22( R R0 )(20 5)2The power dissipated inside the battery isPbatteryE 2 R01.52 5 I R0 0.018 W22( R R0 )(20 5)2One can check thatPbulb Pbattery P12

a simple single-loop circuit. (a) What is the power dissipated in the circuit? (b) What is the power dissipated in the light bulb and in the battery? Light bulb R Solution: (a) The total dissipated power in the circuit is equal to the power of the electromotive force: 0.09 W 20 5

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