Question

Question #088fc

Answer 1

Yes, it will get hotter.

Explanation:

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`

Answer 2

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