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In the Figure above, I was asked to determine which diagram refers to the relation between current frequency and the intensity of the current in an ohmic resistance. I am confused between (2) and (3). Why (2)? Because it is an alternative current, and it is determined from the equation:

i(max)= Vmax/R (R is constant but Vmax=NBA(2πf)) so current would be proprtional to frequency.

Why (3)? Because I thought that in the AC source both voltage and frequency are adjustable as desired, so the change in frequency doesn't imply a change in voltage, so the intensity of current would be stable. But anyway, I am waiting your help to understand what is the correct answer.

  • \$\begingroup\$ To me, it's 3. With a purely resistive load, with no capacitance or inductance, the frequency has no effect whatsoever on the current. I'm assuming that the voltage is constant, since nothing says it isn't. \$\endgroup\$ – Simon B Mar 23 '17 at 23:27
  • \$\begingroup\$ @SimonB, OK, the load is purely resistive and there is no capacitance or inductance, then the frequency will increase and will not affect any thing in the circuit, right? \$\endgroup\$ – Asmaa Mar 24 '17 at 0:07
  • \$\begingroup\$ Your quoted equation is clearly a version of Faraday's Law, so it applies to magnetic coils. An "ohmic resistance" does not, by definition, apply to magnetic coils. ("Ohmic resistance" is in quotes because resistance only applies to DC and to AC which is frequency-independent. The behavior of components whose V/i changes with frequency is called impedance.) So 2 does not apply. 3 is the proper answer. \$\endgroup\$ – WhatRoughBeast Mar 24 '17 at 3:26
  • \$\begingroup\$ @Asmaa If it's a pure ohmic resistance, then the current is given by the RMS voltage and the resistance: I = V/R. The frequency doesn't come into it. In the real World, there will be stray capacitances and inductances, but this is a theoretical question. \$\endgroup\$ – Simon B Mar 24 '17 at 13:34

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