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THEORY QUESTION ON AC CIRCIUT 1. Define the following: I. Reactance II. Impedance Solution: I. Reactance is the opposition in ohms to the alternating current by an inductor, a capacitance or both. Reactance in an AC circuit X, is the ratio of the peak voltage Vo to the peak current Io through a capacitor or an inductor. It is measured in ohms (Ω). II. Impedance is the total opposition in ohms to the alternating current by a combination of resistor as well as a capacitor or an inductor. Impedance in an AC circuit Z is the ratio of the peak voltage Vo to the peak current Io through an AC circuit containing the reactances (either capacitive or inductive, or both) and resistance. 2. Explain resonant frequency of an RLC Circuit. Solution: Resonant frequency f o is the tendency at which the current in the circuit has a maximum value f o 1 2 LC 3. Explain this statement - the power supply voltage of a source is 230V. Solution: It means that the root mean square (r.m.s) value of a voltage is 230V 4. A source of Emf 240V & a frequency of 50Hz are connected to a series arrangement of a resistor, an inductor & a capacitor. When the current in the capacitor is 10A, the potential difference across the resistor is 140V & that across the inductor is 50V. Calculate (i.) The potential difference across the capacitor; (ii.) capacitance of the capacitor; (iii.) inductance of the inductor; (iv.) draw & label one vector diagram for the potential difference across 1|P a g e the inductor, the capacitor & the resistor. XC II . ? 1 wC 1 2fC V C IC Solution: XC Emf = 240V; f = 50Hz; VR 140V ; VL 50V ; I C 10 A ; VC ? ; XC X L ?; X C ?; L ?; C ? Note that same current passes through RLC circuit. Therefore, I C I R I L 10 A . I. VC can be gotten by using Emf 2 VR2 (VL VC ) 2 2402 1402 (50 VC ) 2 57600 19600 (50 VC ) 2 57600 19600 50 VC 38000 50 VC VC 194.95 50 VC 144.95 VC 145V 2|P a g e XC VC 1 IC 2fC C IC 2fVC 10 ( 2)(3.14)(50)(144.95) 10 C 45514.3 C 0.0002197 F C C 219.7 10 6 F C 219.7 F C 220 F XL III . wL X L 2fL XL VL IL VL 2fL IL L VL 2fI L 50 ( 2)(3.14)(50)(10) 50 0.0159 H L 3140 L 15.9mH L 16mH L the capacitor; (ii.) The capacitance of the capacitor; (iii.) The inductance of the inductor. [Take ∏ = 3.14] Solution: Emf = V = 110V; f = 60Hz; IC = 2A; VR = 80V; VC = ?; C=? VL = 40V XL = ? ; XC = ? L =?; Take note that the same current passes through the three devices in the RLC circuit. The diagram is below: IV. VL VECTOR DIAGRAM C R VR VL VR VL V VC I. 5. A source of emf 110V & frequency 60Hz is connected to a resistor, an inductor & a capacitor in series. When the current in the capacitor is 2A, the potential difference across the resistor is 80V & inductor is 40V. Draw the diagram of the potential differences across the inductor, capacitor & the resistor. Calculate (i.) The potential difference across 3|P a g e V 2 VR (VL VC ) 2 2 110 2 80 2 (40 VC ) 2 110 2 80 2 (40 VC ) 2 5700 (40 VC ) 2 Square root 5700 40 VC 75.498 40 VC VC 35.498V | VC | 35.5V VC II. XC VC 1 I C 2fC IC C 2fVC C 2 (2)(3.14)(60)(35.498) 2 C 13375.6464 C 0.0001495F 6 C 1149.5 10 F C 150.0F 6. In a series RC circuit, the resistance of the resistor is 4Ω, the capacitive reactance is 3Ω. Determine the impedance of the circuit. Solution: R = 4Ω; Z 2 R 2 X C2 Z 2 4 2 32 Z 2 16 9 Z 2 25 Z 25 Z 5 III. XL VL wL 2fL IL VL 2fL IL L VL 2fI L 40 ( 2)(3.14)( 60)( 2) 40 L 753.6 L 0.05307856 H L L 53.1 10 3 H L 53.1mH 4|P a g e XC = 3Ω 7. The current I in an AC circuit is given by the equation I 30Sin100t where t is the time in seconds. Deduce the following from the equation: I. Frequency of the current II. Peak value of the current III. RMS value of the current. Solution: I 30 Sin100t Equate with I I 0 Sinwt Where... I 0 30 A; w 100 I. w 2f 100 f 100 2 f 50 Hz II . I 0 30 A I rms III . I 0 I rms 30 I rms 2 2 21.21A Compiled and edited by: FABIYI Clement O. [email protected] http://facebook.com/sum might http://physicsmt.wordpre ss.com © 2014 5|P a g e