Capacitors in Circuits

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## Capacitors in Circuits

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**-Q**A V E d +Q Capacitance Two parallel plates charged Q and –Q respectively constitute a capacitor C = Q / V The relationship C = Q / V is valid for any charge configuration (Indeed this is the definition of capacitance or electric capacity) In the particular case of a parallel plate capacitor C = 0 A / d [vacuum] or C = 0 A / d [dielectric] The capacitance is directly proportional to the area of the plates and inversely proportional to the separation between the plates**Capacitors in Circuits**(Symbol for a capacitor) +Q -Q C V A piece of metal in equilibrium has a constant value of potential. Thus, the potential of a plate and attached wire is the same. The potential difference between the ends of the wires is V, the same as the potential difference between the plates.**Parallel and Series**Series Parallel**Capacitors in Parallel**C1 - q1 • Suppose there is a potential • difference V between a and b. • Then q1 V = C1 & q2 V = C2 a b C2 - q2 • We want to replace C1 and C2 with an • equivalent capacitance C = q V • The charge on C is q = q1 + q2 • Then C = q V = (q1 + q2 ) V = q1 V + q2 V = C1 + C2 V b a C - q C = C1 + C2 • This is the equation for capacitors inparallel. • Increasing the number of capacitors increases the capacitance.**Capacitors in Series**C1 C2 C a -q +q -q +q b a -q +q b V1 V2 V • Here the total potential difference between a and b is V = V1 + V2 • Also V1 = (1/C1) q and V2 = (1/C2) q • The charge on every plate (C1 and C2) must be the same (in magnitude) • Then: V = V1 + V2 = q / C1 +q / C2 = [(1/C1) + (1/C2)] q • or, V = (1/C) q 1 / C = 1 / C1 + 1 / C2 • This is the equation for capacitors in series. • Increasing the number of capacitors decreases the capacitance.**Series**Parallel Ceq = C1 + C2 + C3 1/Ceq = 1/C1 + 1/C2 + 1/C3**Real circuit**Ideal circuit What happens when the switch is closed ? How does the capacitor acquire the charge ?**open**closed I R R VR=IR + + +++ V V - - - VC=q/C - C - C RC Circuits: Charging V = I(t)R + q(t)/C When the switch closes, at first a high current flows: VR is big and VC is small. As q is stored in C, VC increases. This fights against the battery, so I gradually decreases. Finally, I stops (I = 0), C is fully charged (VC = Q/C = V), and Q=C V**R**R VR=IR +q +q I C VC=V0 C -q VC=q/C -q Open circuit After closing switch Discharging an RC Circuit Current will flow through the resistor for a while. Eventually, the capacitor will lose all its charge, and the current will go to zero. During the transient: q(t) / C – I(t) R = 0**Charging and Discharging a Capacitor**Charging and discharging of a capacitor occurs gradually with a characteristic time = RC time constant At t = 0, (switch closed or open) a large current flows, the capacitor behaves like a short circuit. At t , the current is essentially zero, the capacitor behaves like an open switch. The current decreases exponentially.**Measuring**Current and Voltage in a circuit The ammeter measures current, and is connected in series. The voltmeter measures voltage, and is connected in parallel.**A modern digital multimeter combines the functions**of ammeter, voltmeter, and ohmmeter. (i.e. can measure current, voltage, and resistance) In addition, modern multimeters can measure capacitance, temperature, and more, and can be connected to computers too…