How does Kirchhoff's Loop Rule apply to an AC generator with a resistor?

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SUMMARY

The discussion focuses on applying Kirchhoff's Loop Rule to an AC generator connected to a resistor. The equation derived is E_m sin(w_d t) = R I, leading to the current as a function of time given by I = E_m sin(w_d t) / R. For the specific case where R = 50 Ohm and EMF_max = 30.0 V, the amplitude of the alternating current is calculated as 0.6 A. The frequency of the EMF is specified as 1.00 kHz, confirming the parameters for the AC circuit analysis.

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  • Understanding of Kirchhoff's Loop Rule
  • Familiarity with AC circuit analysis
  • Knowledge of differential equations
  • Basic concepts of alternating current (AC) and resistance
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Pete_01
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Homework Statement



A resistor of resistance R is connected to an AC
generator with EMF(t)=(EMF_max)sin((w_d)t). (a) Write
Kirchhoff's Loop Rule for this circuit at a particular
instant of time. Is your result a differential equation?
(b) Find the current as a function of time. Now assume
R=50 Ohm and EMF_max = 30.0 V. (c) What is the amplitude
of the resulting alternating current if the frequency of
the EMF is 1.00 kHz?

Homework Equations



V_R = R dQ(t)/dt

The Attempt at a Solution



Ok, so the differential equation I came up with is:
V_G + V_R = 0
E_m sin(w_d t) = R dQ(t)/dt

assuming that's correct how do I find the current as a function of time? Replace dq/dt with i? And after that?
 
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And after that u get equation [tex]E_m sin(w_d t)=R I[/tex], hence [tex]I=\frac {E_m sin(w_d t)}{R}[/tex] and the amplitude of the ac current is [tex]\frac{E_m}{R}[/tex]

Something else, usually we apply Kirchoff's Law with currents and voltages and not with charges Q. Therefore MAYBE the correct answer to a) is by having replace dq/dt with I from the start and therefore the equation isn't a differential equation.
 
Last edited:
Perfect, thank you!
 

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