SET Chapter 3 PDF
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This document outlines the fundamentals of electric circuits, focusing on Ohm's law. It clearly explains the relationship between voltage, current, and resistance, providing formulas and examples for calculations. The document also includes practice questions to reinforce learning.
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Learning Outcomes Understand the relationship between voltage, current and resistance (Ohm’s law): V=I ×R Apply Ohm’s law to determine voltage, current or resistance in an electric circuit Connect a simple electric circuit comprising a voltmeter, an ammeter, a load and a power supply...
Learning Outcomes Understand the relationship between voltage, current and resistance (Ohm’s law): V=I ×R Apply Ohm’s law to determine voltage, current or resistance in an electric circuit Connect a simple electric circuit comprising a voltmeter, an ammeter, a load and a power supply to verify Ohm’s law 1.3.1 Ohm’s Discovery The relationship between V, I and R was discovered by scientist, Georg Simon Ohm. This discovery, known as Ohm’s law, is the basic formula used in all electric circuits. Ohm’s law states that the current (I ) flowing through a conductor is directly proportional to the potential difference (V ) applied across its ends, provided the temperature remains constant. ↳ voltage/EMF (Voltage increase current , increase) Fig. 1.3-1: Simple electric circuit The following formula is derived from Ohm’s law: V=I×R I = = where V is voltage in volts (V) I is current in amperes (A) R R is resistance in ohms (Ω) Unit 1.3 | Electric Circuit Laws 10 1.3.2 Worked Examples Fig. 1.3-2: Simple electric circuit (a) Refer to Fig. 1.3-2, determine V if I = 0.5 A and R = 20 Ω. V=I ×R = 0.5 A × 20 Ω = 10 V (b) A load of resistance 500 Ω is connected to a supply of 230 V. What is the current drawn from the supply? V V I = R - 230 V F = 500 Ω = 0.46 A Notes ! M : micro -100 000 louA 100000 = = 0. 001A M : willi = 1000 Int = += 0 005A. kilo +1000 20km = 20x1000 20000 - = k : M : Mega x 100000 0 3 MV. = 0 3 +100000. = 30000 Unit 1.3 | Electric Circuit Laws 11 1.3.3 Practice Questions ⑮D V = [xR V I = it I R R = Fig. 1.3-3: Simple electric circuit (a) Referring to Fig. 1.3-3, determine: (i) V if I = 0.5 A and R = 100 Ω. [50 V] V = IXR = 0. 5/ x100m = (ii) I if V = 110 V and R = 550 Ω. [0.2 A] I (iii) R if V = 230 V and I = 1.2 A. [191.67 Ω] V R = I = 4x1000 (iv) I if V = 24 V and R = 10 kΩ (kilo-ohms). [0.0024 A] = =1 ↳ convert to otims I = 0 0024A.. R = 10ke I = 10 X 1000 - 10000 = 100002 (b) A load with resistance of 200 Ω is connected to a supply of 110 V. What is the current drawn from the supply? R 2001 V [0.55 A] llOV I =? =. =. I - (c) The current flowing in a circuit with a load of resistance 50 Ω is 0.2 A. What is the supply voltage? [10 V] R = 50r I 0. =. 2A V IXR = 50x0 2 V ? = =. ↓V. I Unit 1.3 | Electric Circuit Laws 12