Question #088fc

2 Answers
Sep 30, 2017

Yes, it will get hotter.

Explanation:

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As you can see in the phasor diagram, for a capacitor the current leads by a quarter cycle and for a resistance they both have the same phase.

Hence the Net Voltage will be(by Pythagoras theorem)

V = I sqrt(Xc^2 + R^2)

where I is the current Xc is the capacitative reactance and R is the resistance

Now
Xc = 1//omega C

hence when omega increases Xc decreases and consequently I increases. ( V is constant)

When the current increases, so does the Heat loss per second(P)

P=I^2R

Sep 30, 2017

R and C are in parallel: no change in the current through R as frequency increases.
R and C are in series: current through R increases as frequency increases.

Explanation:

The contribution to total circuit impedance, Z, by the resistance, R, is constant as frequency, f, varies. The contribution to Z by the capacitance, C, varies when frequency, f, varies.

We don't know if they are in parallel or in series, so I will discuss both options.

  • If R and C are in parallel, the question is rather trivial, but since the voltage across the resistor is constant at VS, the current through the resistor will not change with frequency.
  • If R and C are in series, the question is more complex. See discussion below.

The impedance, Z, of the series combination of R and the reactance of the C (called X_C) is given by
Z = R + X_C
Z = R + 1/(2*pi*f*C)

The value of Z is affected by frequency, f, changes because the term 1/(2*pi*f*C) is part of the expression for Z.

Since the frequency, f, is in the denominator of the expression for X_C, as frequency increases,
X_C = 1/(2*pi*f*C) decreases.

The value of Z decreases as frequency increases, since it is the sum of R and X_C.

Because
I = V_(ac)/Z,
and Z is in the denominator, this tells us that current increases when frequency increases.

I hope this helps,
Steve