8866 H1 Physics J2/ D.C. Circuits

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1 7. D.C. CIRCUITS Content Practical circuits Series and parallel arrangements Learning Outcomes Candidates should be able to: (a) (b) (c) (d) (e) recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and Abbreviations (AS, 98) and Signs, Symbols and Systematics (AS, 995). draw and interpret circuit diagrams containing sources, switches, resistors, ammeters, voltmeters, and/or any other type of component referred to in the syllabus. solve problems using the formula for the combined resistance of two or more resistors in series. solve problems using the formula for the combined resistance of two or more resistors in parallel. solve problems involving series and parallel circuits for one source of e.m.f.

2 Types of electric current Direct Current (D.C.): Flow of charges in the circuit is in the same direction all the time, from a higher potential to a lower potential (e.g. current from battery). Alternating Current: Flow of charges in the circuit reverses direction at regular intervals (e.g. current from household mains). lectric circuits consist of circuit components (e.g. batteries, resistors, and switches) connected by conductors (e.g. copper cables). For electric current to flow, the circuit components and conductors must form closed loops. There must also be sources of electrical energy (e.g. batteries) and sinks of electrical energy among the circuit components (e.g. resistors and lamps). (a) recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and Abbreviations (AS, 98) and Signs, Symbols and Systematics (AS, 995). lectrical Circuit Symbols Symbol modifier (a) Recall and use variable value inherent non-linearity preset value Conductors and terminals with connection between conductors no connection between conductors current-carrying conductor open terminals ground/earth single pole single throw (SPST) switch single pole double throw (SPDT) switch double pole double throw (DPDT) switch

3 D.C. sources cell battery (> cell) photovoltaic cell d.c. source variable d.c. source Measuring instruments ammeter and milliammeter voltmeter and millivoltmeter galvanometer (alphabetical symbol, null deflection, current detected in certain direction, current detected in another direction) Resistors fixed resistor variable resistor (rheostat) thermistor light-dependent resistor (LDR) potentiometer (voltage divider) Other circuit components diode photodiode light-emitting diode (LD) 3

fuse heating element indicator, lamp or light source Other symbols found in TYS generator motor loudspeaker thermocouple oscilloscope aerial/antenna electric bell buzzer (b) draw and interpret

4 fuse heating element indicator, lamp or light source Other symbols found in TYS generator motor loudspeaker thermocouple oscilloscope aerial/antenna electric bell buzzer (b) draw and interpret circuit diagrams containing sources, switches, resistors, ammeters, voltmeters, and/or any other type of component referred to in the syllabus. Note that, for a certain electric circuit, there are different ways of drawing its circuit diagram. (b) Draw and interpret Actual circuit Circuit diagram Two other possible circuit diagrams for the above electric circuit are as follows: V R R 4

5 xample : Draw the circuit diagram for the electric circuit shown below. Actual circuit Circuit diagram Conservation of charge (must know in order to solve circuit problems) Given that we are dealing with steady currents (i.e. no accumulation of charge at circuit junctions), the sum of currents entering a circuit junction is equal to the sum of currents leaving it. I + + I I3 + I 4 I5 xample Taking currents entering circuit junction as positive and currents leaving circuit junction as negative, we have: I + I I I I Taking currents leaving circuit junction as positive and currents entering circuit junction as negative, we have: I + I + I I I The given diagrams show wires carrying currents I, I, I 3, and I 4, meeting at a junction. Which of the following diagrams represents the equation I + I I 3 + I 4? A B C D I + I 3 + I 4 + I I + I + I 3 I 4 I + I I 3 + I 4 (correct answer) Impossible, all currents are entering and no current leaving. Conservation of energy (must know in order to solve circuit problems) 5

The algebraic sum of e.m.f. (i.e. sources of electrical energy) is equal to the algebraic sum of p.d. (i.e. sinks of energy) for any closed loop within the circuit.

Solution: Conservation of energy for the loop: + I R + I R I ( R + ) R I R + + R Conservation of energy for

(c) solve problems using the formula for the combined resistance of two or more resistors in series.

6 The algebraic sum of e.m.f. (i.e. sources of electrical energy) is equal to the algebraic sum of p.d. (i.e. sinks of energy) for any closed loop within the circuit. xample 3 Find I and I in terms of,, R, R and R 3. Given that 3.0 V.5 V R R R 3 0 Ω find the values of I and I. Solution: Conservation of energy for the loop: + I R + I R I ( R + ) R I R + + R Conservation of energy for the loop: + IR3 I + R 3 Substituting 3.0 V,.5 V and R R R 3 0 Ω, I 0.3 A and I 0.45 A. (c) solve problems using the formula for the combined resistance of two or more resistors in series. Derivation of the effective resistance of resistors in series (for reference only): 6

7 For series connection, the effective resistance is the sum of individual resistance. In general, for n similar resistors (each of resistance R) connected in series, the effective resistance is R eff nr. Note: The effective resistance always increases when additional resistors are connected in series.. Same water current flowing through each section of the tube, whether wide or narrow (water cannot be compressed, no accumulation, inflow outflow).. Resistance to water flow increases as number of narrow portions along the tube increases (adding obstruction to water flow). xample 4 Calculate the effective resistance of a 4 Ω and two 3 Ω resistors connected in series. Solution: R eff 4 + ()(3) 0 Ω (d) solve problems using the formula for the combined resistance of two or more resistors in parallel. Derivation of the effective resistance of resistors in parallel (for reference only): For parallel connection, the reciprocal of effective resistance is the sum of reciprocal of individual resistance. 7

8 In general, for n similar resistors (each of resistance R) connected in parallel, the n R effective resistance is R eff. R eff R n Note: The effective resistance always decreases when additional resistors are connected in parallel. The effective resistance of resistors in parallel is always less than the individual resistance of each resistor.. The water current through the wide tube is the same as the sum of the water currents in each of the narrow tubes (water cannot be compressed, no accumulation, inflow outflow).. Resistance to water flow decreases as number of narrow tubes increases (adding channels for water flow). xample 5 Calculate the effective resistance of a Ω, a 3 Ω and a 4 Ω resistor connected in parallel. Solution: R eff Reff Ω 3 (e) solve problems involving series and parallel circuits for one source of e.m.f. xample 6 A battery C of.5 V and negligible internal resistance is connected to the combination of resistors as shown. Find the values of the currents I, I, and I 3 in the diagram. 0 Ω 0 Ω I A C I 5 Ω 5 Ω I 3 I 5I 3 I 5I 3 3I I I + I3 I + 3I 4I 8

9 0 Ω I 0.09 I A C 3.75 Ω I I 3 I I A More worked examples Find the equivalent resistances between points A and point B in xamples 7, 8, and 9. Assume that all resistances in the circuits are Ω each. xample 7 B A 9

10 Solution: Step : Assign potential at each terminal of each resistors. V F V D Step : Group resistors which have same potential at its terminals. V F V D Step 3: Simplify the circuit Ω // Ω Ω + Ω Ω.33 Ω Ω //4 Ω.33 Ω + 4 V F Ω V D 0

11 3 Ω // Ω. Ω Ω. Ω V D Hence, effective resistance between point A and point B Ω xample 8 A B Solution: Step : Assign potential at each terminal of each resistors V F V D

12 Step : Group resistors which have same potential at its terminals. V F V D Ω // Ω //4 Ω 0.8 Ω Step 3: Simplify the circuit Ω //( Ω Ω ). Ω +.8 Hence, effective resistance between point A and point B +.. Ω xample 9 A B

13 Solution: Step : Assign potential at each terminal of each resistors Step : Group resistors which have same potential at its terminals. Step 3: Simplify the circuit 0 Ω (short circuit) Hence, effective resistance between point A and point B Ω --ND-- 3

Theme 5: Electricity in the Home

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