Unformatted text preview: Resistive Load Lets first consider an alternating emf hooked up to a resistor. By the loop rule v R = E m sin( ω d t ) = V R sin( ω d t ) We are using the notation still that capital letters denote amplitudes and lower case letters denote time dependent quantities. Hence, the current is i R = v R R = V R R sin( ω d t ) = I R sin( ω d t- φ ) So, the current amplitude is related to the potential amplitude in the usual Ohm’s law way. V R = I R R (45) We can also see by comparison that φ = 0 and therefore the potential and the current in a circuit are in phase. Capacitive Load Now, let us give the same treatment to a capacitor hooked up to a driving voltage source. The voltage across the capacitor is set by the driving emf. v C = V C sin ω d t The charge and current are therefore given by q = Cv C = CV C sin ω d t ⇒ i C = ω d CV C cos ω d t = I C sin( ω d t- φ ) By making comparisons in the second equation above we can see that, for a capacitive load, φ =- 90 ◦ . We can also identify an Ohm’s law type equation for the relation between the current....
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- Spring '08
- Physics, Sin, iL, Electrical resistance, capacitive load