Abstract- The development of De-Sauty’s bridge for capacitance measurement has been followed by various modifications and improvements in the original circuit. Lately Schering bridge and transformer ratio arm bridge have been used for measurement of capacitance in a circuit.Today there is instrumentation available that makes use of these bridges in real life such as that can measure the quality of electrical insulation at power frequency as well as in touch-sensitive sensors. This paper
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capacitors in parallel? Why? 3. You have two capacitors of 0.40 F and 0.60 F. a) What is the equivalent capacitance if you add them in series? b) What is the equivalent capacitance if you add them in parallel? 4. Three capacitors: 0.20 F‚ 0.08 F and 0.40 F‚ are connected in series and attached to a 12 V battery. a) What is the total equivalent capacitance of this arrangement? b) What charge accumulates on each capacitor? c) What is the voltage across each
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Title : C2 Investigating the capacitance of a capacitor I. Objective : * To investigate the factor which affect the capacitance of a parallel-plate capacitor using a reed switch. II. Apparatus * Reed Switch * Signal generator * Capacitor Plates (1 pair) * Polythene spacers * Low voltage power supply * Voltmeter * Variable resistance * Light-beam galvanometer * CRO * Standard mass (eg. 100g ) * Connecting leads * Drawing board *
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Date: 25th March‚ 2010 Title: Investigating the capacitance of a parallel-plate capacitor using a reed switch Objective: To investigate the factors which affect the capacitance of a parallel-plate capacitor using a reed switch. Apparatus: - reed switch - signal generator - capacitor plates of area about 0.24m X 0.24 m 1 pair - polythene spacers ( 10 X 10 X 1 mm ) - polythene sheet‚ same area as capacitor plate 1 mm thick - battery box with 4 cells
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6 Inductance‚ Capacitance‚ and Mutual Inductance Assessment Problems AP 6.1 [a] ig = 8e−300t − 8e−1200t A v=L dig = −9.6e−300t + 38.4e−1200t V‚ dt 38.4e−1200t = 9.6e−300t t > 0+ v(0+ ) = −9.6 + 38.4 = 28.8 V [b] v = 0 when or t = (ln 4)/900 = 1.54 ms [c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W dp = 0 when e1800t − 12.5e900t + 16 = 0 [d] dt Let x = e900t x = 1.45‚ x = 11.05‚ and solve the quadratic x2 − 12.5x + 16 = 0 t= ln 1.45 = 411.05 µs 900 ln 11.05 = 2.67 ms 900 t= p is
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2a Homework 08 - Solutions You need a 45 Ω resistor‚ but the stockroom has only 20 Ω and 50 Ω resistors. How can the desired resistance be achieved under these circumstances?" " A. Put one 50 Ω and one 20 Ω resistor in parallel and then put that combination in series with a 20 Ω resistor." B. Put two 50 Ω resistors in series and then put that combination in parallel with a 20 Ω resistor." C. Put one 50 Ω and one 20 Ω resistor in series and then put that combination in parallel with
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length of time a capacitor took to charge and discharge. We used the average of the charging and discharging time constants to calculate the capacitance by using the equation τ = RC. Since we know the resistance and the time constant‚ we are able to solve for the capacitance and compared the observed and theoretical values in order to verify the capacitance. For the RC circuit with one capacitor‚ we compared the theoretical and observed time constant and obtained a percent difference of 9.5%. For
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Chapter 26 Capacitance and Dielectrics Multiple Choice 1. Determine the equivalent capacitance of the combination shown when C = 12 pF. C C 2C C a. b. c. d. e. 2. 48 pF 12 pF 24 pF 6.0 pF 59 pF Determine the equivalent capacitance of the combination shown when C = 15 mF. C C 2C C a. b. c. d. e. 3. 20 mF 16 mF 12 mF 24 mF 75 mF Determine the equivalent capacitance of the combination shown when C = 12 nF. 2C C C 3C a. b. c. d. e. 34 nF 17 nF 51 nF 68
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EXERCISE Question 1 a) Two point of charges of opposite sign and same magnitude 3.5 μC are separated by a distance of 5 cm. Find (i) the magnitude and direction of the electric field strength‚ E at the midpoint between the charges. (ii) the magnitude and direction of the force F on an electron placed at the midpoint. Answer: i) E net = 1.008x108 N/C toward the –ve charge ii) F net = 1.6128x10-11 N toward +ve charge
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September 27‚ 2011. Date of submission October 30 ‚ 2011. 1 Experiment :B-03 Determination of self inductance of a coil by Rayleigh’s method. Theory : Figure 1: Rayleigh’s method circuit diagram. If a potential V applied to a condenser of capacitance C‚ imparts Q units of charge to the condenser then‚ Q V From the ballistic galvanometer working formula we got: C= T i θ1 θλ (1 + ) πθ 2 2 T i θ1 θ1 Q= π θs 2 θ3 Where‚ θs = the maximum displacement of the steady deflection‚ T=period of the oscillation
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