Electricity Part 4
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Chapter 25 : Electric circuits • Voltage and current • Series and parallel circuits • Resistors and capacitors • Kirchoff’s rules for analysing circuits Electric circuits • Closed loop of electrical components around which current can flow, driven by a potential difference • Current (in Amperes A) is the rate of flow of charge • Potential difference (in volts V) is the work done on charge Electric circuits • May be represented by a circuit diagram. Here is a simple case: • R is the resistance (in Ohms Ω) to current flow Electric circuits • Same principles apply in more complicated cases! Electric circuits • How do we deal with a more complicated case? What is the current flowing from the battery? Electric circuits • When components are connected in series, the same electric current flows through them πΌπΌ πΌπΌ • Charge conservation : current cannot disappear! Electric circuits • When components are connected in parallel, the same potential difference drops across them ππ ππ ππ • Points connected by a wire are at the same voltage! Electric circuits • When there is a junction in the circuit, the inward and outward currents to the junction are the same πΌπΌ1 πΌπΌ2 πΌπΌ3 πΌπΌ1 = πΌπΌ2 + πΌπΌ3 • Charge conservation : current cannot disappear! Consider the currents I1, I2 and I3 as indicated on the circuit diagram. If I1 = 2.5 A and I2 = 4 A, what is the value of I3? 1. 2. 3. 4. 5. 6.5 A 1.5 A −1.5 A 0A The situation is not possible I1 I2 0% 1 0% 2 0% 3 I3 0% 4 0% 5 Consider the currents I1, I2 and I3 as indicated on the circuit diagram. If I1 = 2.5 A and I2 = 4 A, what is the value of I3? I2 I1 current in = current out πΌπΌ1 = πΌπΌ2 + πΌπΌ3 πΌπΌ3 = πΌπΌ1 − πΌπΌ2 πΌπΌ3 = 2.5 − 4 = −1.5 π΄π΄ (Negative sign means opposite direction to arrow.) I3 A 9.0 V battery is connected to a 3 β¦ resistor. Which is the incorrect statement about potential differences (voltages)? 1. 2. 3. 4. Vb – Va = 9.0 V Vb – Vc = 0 V Vc – Vd = 9.0 V Vd – Va = 9.0 V a b d c 0% 1 0% 2 0% 3 0% 4 Resistors in circuits • Resistors are the basic components of a circuit that determine current flow : Ohm’s law I = V/R Resistors in series/parallel • If two resistors are connected in series, what is the total resistance? same current π π 1 πΌπΌ Potential drop ππ1 = πΌπΌ π π 1 π π 2 πΌπΌ Potential drop ππ2 = πΌπΌ π π 2 Total potential drop ππ = ππ1 + ππ2 = πΌπΌ π π 1 + πΌπΌ π π 2 = πΌπΌ (π π 1 + π π 2 ) Resistors in series/parallel • If two resistors are connected in series, what is the total resistance? π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ πΌπΌ Potential drop ππ = πΌπΌ π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = πΌπΌ (π π 1 + π π 2 ) π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = π π 1 + π π 2 • Total resistance increases in series! Resistors in series/parallel • Total resistance increases in series! Resistors in series/parallel • If two resistors are connected in parallel, what is the total resistance? π π 1 π π 2 Resistors in series/parallel • If two resistors are connected in parallel, what is the total resistance? πΌπΌ πΌπΌ1 πΌπΌ2 π π 1 π π 2 ππ Total current πΌπΌ = πΌπΌ1 + πΌπΌ2 = πΌπΌ ππ π π 1 + ππ π π 2 = ππ 1 π π 1 1 + π π 2 Resistors in series/parallel • If two resistors are connected in parallel, what is the total resistance? π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ Current I = ππ π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = ππ 1 π π 1 + 1 π π 2 πΌπΌ 1 π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ • Total resistance decreases in parallel! 1 1 = + π π 1 π π 2 Resistors in series/parallel • Total resistance decreases in parallel! Resistors in series/parallel • What’s the current flowing? (1) Combine these 2 resistors in parallel: 1 π π ππππππππ 1 1 + = 30 50 π π ππππππππ = 18.75 Ω (2) Combine all the resistors in series: π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = 20 + 18.75 + 20 = 58.75 Ω (3) Current πΌπΌ = ππ π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = 10 58.75 = 0.17 π΄π΄ If an additional resistor, R2, is added in series to the circuit, what happens to the power dissipated by R1? R1 1. Increases 2. Decreases 3. Stays the same ππ = πΌπΌπΌπΌ 0% 1 0% 0% 2 2 ππ ππ = ππππ = πΌπΌ 2 π π = π π 3 If an additional resistor, R3, is added in parallel to the circuit, what happens to the total current, I? 1. 2. 3. 4. I V Increases Decreases Stays the same Depends on R values Parallel resistors: reciprocal effective resistance is sum of reciprocal resistances 0% 1 1 π π ππππππ 0% 2 0% 3 0% 4 1 1 1 = + + +β― π π 1 π π 2 π π 3 Series vs. Parallel CURRENT VOLTAGE Same current through all series elements Voltages add to total circuit voltage RESISTANCE Adding resistance increases total R Current “splits up” through parallel branches Same voltage across all parallel branches Adding resistance reduces total R String of Christmas lights – connected in series Power outlets in house – connected in parallel Voltage divider Consider a circuit with several resistors in series with a battery. Current in circuit: πΌπΌ = ππ π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ ππ = π π 1 + π π 2 + π π 3 The potential difference across one of the resistors (e.g. R1) ππ1 = πΌπΌπ π 1 π π 1 = ππ π π 1 + π π 2 + π π 3 The fraction of the total voltage that appears across a resistor in series is the ratio of the given resistance to the total resistance. What must be the resistance R1 so that V1 = 2.0 V? ππ1 R1 12 V 6.0 β¦ 1. 0.80 β¦ 2. 1.2 β¦ 3. 6.0 β¦ 4. 30 β¦ 0% 1 0% 2 0% 3 0% 4 Capacitors • A capacitor is a device in a circuit which can be used to store charge A capacitor consists of two charged plates … Electric field E It’s charged by connecting it to a battery … Capacitors • A capacitor is a device in a circuit which can be used to store charge Example : store and release energy … Capacitors • The capacitance C measures the amount of charge Q which can be stored for given potential difference V +Q -Q ππ πΆπΆ = ππ ππ = πΆπΆ ππ (Value of C depends on geometry…) V • Unit of capacitance is Farads [F] Resistor-capacitor circuit • Consider the following circuit with a resistor and a capacitor in series V What happens when we connect the circuit? Resistor-capacitor circuit • When the switch is connected, the battery charges up the capacitor • Move the switch to point a • Initial current flow I=V/R V • Charge Q flows from battery onto the capacitor • Potential across the capacitor VC=Q/C increases • Potential across the resistor VR decreases • Current decreases to zero Resistor-capacitor circuit • When the switch is connected, the battery charges up the capacitor V Resistor-capacitor circuit • When the battery is disconnected, the capacitor pushes charge around the circuit • Move the switch to point b • Initial current flow I=VC/R V • Charge flows from one plate of capacitor to other • Potential across the capacitor VC=Q/C decreases • Current decreases to zero Resistor-capacitor circuit • When the battery is disconnected, the capacitor pushes charge around the circuit V Capacitors in series/parallel • If two capacitors are connected in series, what is the total capacitance? Same charge must be on every plate! +ππ -ππ πΆπΆ1 Potential drop ππ1 = ππ/πΆπΆ1 +ππ πΆπΆ2 -ππ ππ = πΆπΆ ππ Potential drop ππ2 = ππ/πΆπΆ2 Total potential drop ππ = ππ1 + ππ2 = ππ πΆπΆ1 + ππ πΆπΆ2 = ππ 1 πΆπΆ1 + 1 πΆπΆ2 Capacitors in series/parallel • If two capacitors are connected in series, what is the total capacitance? -ππ +ππ Potential drop ππ = ππ πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = ππ 1 πΆπΆ1 + 1 πΆπΆ2 ππ = πΆπΆ ππ 1 πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ • Total capacitance decreases in series! 1 1 = + πΆπΆ1 πΆπΆ2 Capacitors in series/parallel • If two capacitors are connected in parallel, what is the total capacitance? +ππ1 +ππ2 ππ πΆπΆ1 πΆπΆ2 −ππ1 −ππ2 ππ = πΆπΆ ππ Total charge ππ = ππ1 + ππ2 = πΆπΆ1 ππ + πΆπΆ2 ππ = πΆπΆ1 + πΆπΆ2 ππ Capacitors in series/parallel • If two capacitors are connected in parallel, what is the total capacitance? +ππ -ππ πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ Total charge ππ = πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ ππ = πΆπΆ1 + πΆπΆ2 ππ ππ = πΆπΆ ππ πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = πΆπΆ1 + πΆπΆ2 • Total capacitance increases in parallel! Two 5.0 F capacitors are in series with each other and a 1.0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 1. 2. 3. 4. Q = 5.0 C, Qtotal = 5.0 C Q = 0.25 C, Qtotal = 0.50 C Q = 2.5 C, Qtotal = 2.5 C Q = 2.5 C, Qtotal = 5.0 C ππ = πΆπΆπΆπΆ 5F 1V 0% 1. 1 πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ 5F 0% 2. 0% 3. 0% 4. 1 1 1 = + + +β― πΆπΆ1 πΆπΆ2 πΆπΆ3 Two 5.0 F capacitors are in series with each other and a 1.0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 5F 1V 5F Potential difference across each capacitor = 0.5 V 1 Charge on each capacitor ππ = πΆπΆπΆπΆ = 5 × 0.5 = 2.5 πΆπΆ πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ 1 1 1 1 1 2 → πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = 2.5 πΉπΉ = + → = + = πΆπΆ1 πΆπΆ2 πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ 5 5 5 πππ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ ππ = 2.5 × 1 = 2.5 πΆπΆ Two 5.0 F capacitors are in parallel with each other and a 1.0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 1. 2. 3. 4. Q = 5.0 C, Qtotal = 5.0 C Q = 0.2 C, Qtotal = 0.4 C Q = 5.0 C, Qtotal = 10 C Q = 2.5 C, Qtotal = 2.5 C ππ = πΆπΆπΆπΆ 1V 0% 1. 5F 0% 2. 0% 3. 5F 0% 4. πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = πΆπΆ1 + πΆπΆ2 + πΆπΆ3 + β― Two 5.0 F capacitors are in parallel with each other and a 1.0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 1V Potential difference across each capacitor = 1 V Charge on each capacitor ππ = πΆπΆπΆπΆ = 5 × 1 = 5 πΆπΆ πππ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = 10 πΆπΆ 5F 5F Resistors vs. Capacitors π π 1 πΆπΆ1 π π 2 πΆπΆ2 π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = π π 1 + π π 2 1 πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ 1 1 = + πΆπΆ1 πΆπΆ2 Resistors vs. Capacitors π π 1 1 1 1 = + π π 1 π π 2 π π 2 π π π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ πΆπΆ1 πΆπΆπ‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ = πΆπΆ1 + πΆπΆ2 πΆπΆ2 Kirchoff’s rules • Sometimes we might need to analyse more complicated circuits, for example … π π 1 = 4 Ω ππ1 = 9 ππ π π 2 = 1 Ω ππ2 = 6 ππ π π 3 = 2 Ω Q) What are the currents flowing in the 3 resistors? • Kirchoff’s rules give us a systematic method Kirchoff’s rules • What are the currents flowing in the 3 resistors? 9 ππ 6 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 2Ω πΌπΌ3 Kirchoff’s junction rule : the sum of currents at any junction is zero Kirchoff’s rules • The sum of currents at any junction is zero • Watch out for directions : into a junction is positive, out of a junction is negative πΌπΌ1 πΌπΌ2 πΌπΌ3 πΌπΌ1 − πΌπΌ2 − πΌπΌ3 = 0 πΌπΌ1 = πΌπΌ2 + πΌπΌ3 Kirchoff’s rules • What are the currents flowing in the 3 resistors? 9 ππ 6 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 2Ω πΌπΌ3 Kirchoff’s junction rule : the sum of currents at any junction is zero πΌπΌ2 πΌπΌ1 πΌπΌ3 πΌπΌ1 + πΌπΌ2 + πΌπΌ3 = 0 Kirchoff’s rules • What are the currents flowing in the 3 resistors? 9 ππ 6 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 2Ω πΌπΌ3 Kirchoff’s loop rule : the sum of voltage changes around a closed loop is zero 9 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 Kirchoff’s rules • Sum of voltage changes around a closed loop is zero • Consider a unit charge (Q=1 Coulomb) going around this loop • It gains energy from the battery (voltage change +V) • It loses energy in the resistor (voltage change - I R) • Conservation of energy : V - I R = 0 (or as we know, V = I R) Kirchoff’s rules • What are the currents flowing in the 3 resistors? 9 ππ 6 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 2Ω πΌπΌ3 Kirchoff’s loop rule : the sum of voltage changes around a closed loop is zero 9 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 9 − 4 πΌπΌ1 + 1 πΌπΌ2 = 0 Kirchoff’s rules • What are the currents flowing in the 3 resistors? 9 ππ 6 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 2Ω πΌπΌ3 Kirchoff’s loop rule : the sum of voltage changes around a closed loop is zero −6 − 1 πΌπΌ2 + 2 πΌπΌ3 = 0 Kirchoff’s rules • What are the currents flowing in the 3 resistors? 9 ππ 6 ππ 4Ω πΌπΌ1 1Ω πΌπΌ2 2Ω πΌπΌ3 We now have 3 equations: πΌπΌ1 + πΌπΌ2 + πΌπΌ3 = 0 (1) 9 − 4πΌπΌ1 + πΌπΌ2 = 0 (2) −6 − πΌπΌ2 + 2πΌπΌ3 = 0 (3) To solve for πΌπΌ1 we can use algebra to eliminate πΌπΌ2 and πΌπΌ3 : 1 → πΌπΌ3 = −πΌπΌ1 − πΌπΌ2 ππππππ. ππππ 3 → πΌπΌ2 = −2 − 23 πΌπΌ1 ππππππ. ππππ 2 → πΌπΌ1 = 1.5 π΄π΄ Consider the loop shown in the circuit. The correct Kirchoff loop equation, starting at “a” is 1. π°π°ππ πΉπΉππ + πΊπΊππ + π°π°ππ πΉπΉππ + πΊπΊππ + π°π°ππ πΉπΉππ = ππ 2. −π°π°ππ πΉπΉππ − πΊπΊππ − π°π°ππ πΉπΉππ − πΊπΊππ − π°π°ππ πΉπΉππ = ππ 3. π°π°ππ πΉπΉππ − πΊπΊππ + π°π°ππ πΉπΉππ − πΊπΊππ − π°π°ππ πΉπΉππ = ππ 4. π°π°ππ πΉπΉππ − πΊπΊππ + π°π°ππ πΉπΉππ + πΊπΊππ + π°π°ππ πΉπΉππ = ππ 0% 1 0% 2 0% 3 0% 4 Chapter 25 summary • Components in a series circuit all carry the same current • Components in a parallel circuit all experience the same potential difference • Capacitors are parallel plates which store equal & opposite charge Q = C V • Kirchoff’s junction rule and loop rule provide a systematic method for analysing circuits
