Question

In one circuit, a resistance R and inductance L are connected to an AC voltage source with variable voltage V = √2 V0 cos(ωt)]. A second circuit, with the same resistance and inductance, has a DC voltage source V0. They have been connected for a long time.

(a) Show that when L = 0, the average power through the resistor in the AC circuit is the same as the power through the resistor in the DC circuit.
(b) What is the peak current in the AC circuit in terms of L, R, ω and V0?
(c) In terms of R and L, what frequency ω would produce average power in the resistor in the AC circuit compared to the power in the DC circuit?

Answer 1

Step-by-step explanation:

Given: i. in the AC series circuit, V = √2
`V_(0)`cos(ωt)], L and R.

ii. in the DC circuit, R, L and
`V_(0)`.

1. To show that average power through resistor in the AC circuit is the same in the DC circuit, when L = 0.

For the AC circuit,

Average power =
`I^(2)`Z

where: I is the current flowing and Z is the impedance.

Impedance, Z =
`\sqrt{R^(2) + X_(L) ^(2) }`

But,
`X_(L)` = 2
`\pi`fL

⇒
`X_(L)` = 0 (∵ L = 0)

So that, Impedance, Z = R

Thus, average power through the resistor in the AC circuit =
`I^(2)`R

For the DC circuit,

Average power =
`I^(2)`R

Therefore,

the average power through the resistor in the AC circuit = the average power through the resistor in the DC circuit =
`I^(2)`R, when L = 0.

2. In the AC circuit, the expression for the peak current,
`I_(0)` can be determined by:

`V_(0)` =
`I_(0)` × Z

`I_(0)` =
`(V_(0) )/(Z)`

`I_(0)` =
`\frac{V_(0) }{\sqrt{R^(2) + (wL)^(2) } }`

3. The frequency that would produce average power in the AC circuit is;

Average power =
`I^(2)`Z

Average power =
`I^(2)`(
`\sqrt{R^(2) + (wL)^(2) }`

Square both sides to have;

`(Average power)^(2)` =
`I^(4)` (
`R^(2)` +
`w^(2)`
`L^(2)`)

`(Average power)^(2)` -
`I^(4)`
`R^(2)` =
`w^(2)`
`L^(2)`

⇒
`w^(2)` = [
`(Average power)^(2)` -
`I^(4)`
`R^(2)`] ÷
`L^(2)`

ω =
`\sqrt{}`{[
`(Average power)^(2)` -
`I^(4)`
`R^(2)`] ÷
`L^(2)`}

Therefore, frequency, ω =
`\sqrt{}`{[
`(Average power)^(2)` -
`I^(4)`
`R^(2)`] ÷
`L^(2)`} is required to produce average power in the resistor in the AC circuit compared to the DC circuit.