Alternating Currents

(d) $\omega = \dfrac{1}{LC}$

At resonance $X_L = X_C$, so $L\omega = \dfrac{1}{C\omega}$. This gives $\omega^2 = \dfrac{1}{LC}$, so $\omega = \dfrac{1}{\sqrt{LC}}$. Relations (a), (b) and (c) are all forms of this result. Relation (d) is missing the square root, so it does not hold.

(a) Impedance of a series LCR circuit

Setup: A resistor $R$, an inductor $L$ and a capacitor $C$ are in series with an ac source $V = V_0\sin\omega t$. The same current $I = I_0\sin\omega t$ flows through all three, so we take the current as the reference phasor.

     R          L          C
  --/\/\/---(((((((---| |---
  |                        |
  -----------( ~ )----------

1. The voltage across each element is:
   • $V_R = IR$, in phase with $I$
   • $V_L = IX_L$ with $X_L = \omega L$, leading $I$ by $90^\circ$
   • $V_C = IX_C$ with $X_C = \dfrac{1}{\omega C}$, lagging $I$ by $90^\circ$
2. $V_L$ and $V_C$ point in opposite directions. Take $X_L > X_C$; their resultant is $(V_L - V_C)$, at $90^\circ$ to $V_R$.

        ^ V_L - V_C
        |        / V
        |      /
        |    /  phi
        |  /
        +--------------> I, V_R

3. Add the phasors $V_R$ and $(V_L - V_C)$:

4. So

5. Impedance is defined by $Z = \dfrac{V}{I}$:

The phase angle is $\tan\phi = \dfrac{X_L - X_C}{R}$.

(b) Condition of resonance

The current $I = \dfrac{V}{Z}$ is largest when $Z$ is smallest. This happens when the reactive part becomes zero:

At resonance $Z = R$ (its minimum value), the current is maximum ($I = V/R$), the current is in phase with the voltage, and the power factor is 1.

(c) Numerical (RC circuit)

Given: $C = 2\ \mu\text{F} = 2\times10^{-6}$ F, $R = 1000\ \Omega$, $V_{rms} = 12$ V, $f = 50$ Hz

1. Capacitive reactance:

2. Impedance:

3. Current:

4. Voltage across the capacitor:

5. Phase angle:

Result: $I \approx 6.4$ mA (rms), $V_C \approx 10.2$ V, and the current leads the applied emf by $\phi \approx 58^\circ$.

(a) $\dfrac{\pi}{3}$

The power factor is $\cos\phi$. Here $\cos\phi = 0.5$, so $\phi = 60^\circ = \dfrac{\pi}{3}$.

(a) Relation between current and voltage

Setup: $R$, $L$ and $C$ are in series with an ac source $V = V_0\sin\omega t$. The same current $I = I_0\sin\omega t$ flows through every element.

1. The voltages across the three elements are:
   • $V_R = IR$, in phase with $I$
   • $V_L = IX_L$ ($X_L = \omega L$), leading $I$ by $\pi/2$
   • $V_C = IX_C$ ($X_C = 1/\omega C$), lagging $I$ by $\pi/2$
2. $V_L$ and $V_C$ are opposite in direction, so their resultant is $(V_L - V_C)$, perpendicular to $V_R$.
3. The applied voltage is the phasor sum of these:

4. Therefore

5. The voltage and current differ in phase by $\phi$, where

So $V = V_0\sin\omega t$ gives $I = \dfrac{V_0}{Z}\sin(\omega t - \phi)$.

(b) Phasor diagram

  Case X_L > X_C (voltage leads current by phi)

        ^ V_L
        |
        |  ^ (V_L - V_C)     V
        |  |              /
        |  |           /
        |  |        /  phi
        |  |     /
        +--+--------------> V_R  (along I)
        |
        v V_C

• If $X_L > X_C$, the circuit is inductive and the voltage leads the current by $\phi$.
• If $X_C > X_L$, the resultant $(V_C - V_L)$ points downward and the current leads the voltage by $\phi$.
• If $X_L = X_C$ (resonance), $\phi = 0$ and the voltage and current are in phase.

(c) Numerical

Given: $R = 25\ \Omega$, $L = 30$ mH $= 0.030$ H, $C = 10\ \mu\text{F} = 1.0\times10^{-5}$ F, $V = 240$ V, $f = 50$ Hz

1. Inductive reactance:

2. Capacitive reactance:

3. Impedance:

4. Current:

5. Voltmeter reading across the capacitor:

Result: current $\approx 0.77$ A; voltmeter across the capacitor reads $\approx 246$ V. This is more than the supply voltage, which is possible because $V_C$ and $V_L$ are out of phase and partly cancel each other.

a) Impedance

Impedance $Z$ of an LCR circuit is the total effective opposition that the circuit offers to the flow of alternating current. It is the ratio $V/I$ (rms or peak values), and

Its unit is the ohm ($\Omega$).

b) i) Impedance of the given circuit

Given (from the figure): $L = 60$ mH $= 0.060$ H, $C = 0.50\ \mu\text{F} = 5.0\times10^{-7}$ F, $R = 300\ \Omega$, $f = 50$ Hz

b) ii) Resonance

Resonance occurs when the inductive reactance equals the capacitive reactance, $X_L = X_C$. Then $Z = R$ (minimum), the current is maximum, and the current is in phase with the voltage.

For this circuit:

So at 50 Hz the circuit is not at resonance. It would resonate if the source frequency were raised to about 919 Hz.

c) i) Identifying device X

The current $I = I_0\sin\left(\omega t + \dfrac{\pi}{2}\right)$ leads the voltage by $\dfrac{\pi}{2}$. This happens only in a purely capacitive circuit, so X is a capacitor.

Its reactance is

c) ii) Graphs over one cycle

 V, I        v = voltage V0 sin(wt),  i = current I0 sin(wt + pi/2)
  ^
  |iiii     vvvvvvv                             iiii
  |    iiivv       vvv                       iii
  |   vvv iii         vvv                 iii
  | vv       ii          vv             ii
  +v-----------i-----------v-----------i-----------v--> t
  |             ii          vv       ii          vv
  |               iii         vvv iii         vvv
  |                  iii       iiivv       vvv
  |                     iiiiiii     vvvvvvv
  0           T/4         T/2         3T/4        T

The current curve is a cosine curve. It reaches its peak a quarter period ($T/4$) before the voltage curve, which shows that the current leads by $90^\circ$.

(C) 310 V

The 220 V supply is an rms value, so $V_0 = \sqrt2\,V_{rms} = 1.414\times220 \approx 311$ V, which is about 310 V.

i) Phase angle

Given: $L = 100$ mH $= 0.1$ H, $C = 2\ \mu\text{F} = 2\times10^{-6}$ F, $R = 400\ \Omega$, $f = \dfrac{500}{\pi}$ Hz

1. Angular frequency:

2. Reactances:

3. Phase angle:

The voltage and current differ in phase by $45^\circ$.

ii) Which leads?

The current leads the voltage, by $45^\circ$. Here $X_C$ (500 $\Omega$) is greater than $X_L$ (100 $\Omega$), so the circuit is capacitive. The negative value of $\phi$ shows that the voltage lags behind the current.

Impedance $Z$ of an ac circuit is the total effective opposition offered by the circuit (resistance and reactances together) to the flow of alternating current. It is the ratio of rms voltage to rms current, $Z = \dfrac{V_{rms}}{I_{rms}}$. For a series LCR circuit $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Its unit is the ohm ($\Omega$).

(c) 10 $\Omega$

For a coil with resistance and inductance, $Z = \sqrt{R^2 + X_L^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \Omega$.

a) rms value of current

The rms (root mean square) value of an alternating current is the steady (dc) current that would produce the same heat in a given resistor in the same time as the ac does. It equals the square root of the mean of $I^2$ over one cycle:

b) Voltage leads current by $\pi/2$ in a pure inductor

Setup: An emf $E = E_0\sin\omega t$ is applied across a pure inductor $L$ (no resistance).

       L
  ---(((((((---
  |           |
  ----( ~ )----
     E = E0 sin wt

1. The applied emf is balanced by the back emf, so

2. Integrate with respect to time:

3. Since $-\cos\omega t = \sin\left(\omega t - \dfrac{\pi}{2}\right)$,

4. Compare with $E = E_0\sin\omega t$. The current lags behind the emf by $\dfrac{\pi}{2}$; in other words, the voltage leads the current by $\dfrac{\pi}{2}$. The inductive reactance is $X_L = \omega L$.

Phasor diagram:

        ^  E0  (voltage)
        |
        |
        | 90 deg
        +----------> I0  (current)

The voltage phasor is $90^\circ$ ahead of the current phasor. Both rotate anticlockwise with angular velocity $\omega$.

c) Why a capacitor blocks dc

The capacitive reactance is $X_C = \dfrac{1}{2\pi fC}$. For dc, $f = 0$, so $X_C \to \infty$. Once the capacitor is charged to the supply voltage, no further current flows, because the plates are separated by an insulator. So a capacitor blocks dc but allows ac to pass.

(a) Root mean square value

The rms value of an alternating current is the value of steady direct current which produces the same heating effect in a resistor in a given time. It is the square root of the mean of the squared current over a cycle: $I_{\text{rms}} = \sqrt{\overline{I^2}} = \dfrac{I_0}{\sqrt2}$.

(b)(i) Impedance

Let the current be $I = I_0\sin\omega t$. In series the same current flows in all elements, and the voltage drops are

• across $R$: $V_R = IR$, in phase with $I$;
• across $L$: $V_L = IX_L$, leading $I$ by $90^\circ$ ($X_L = \omega L$);
• across $C$: $V_C = IX_C$, lagging $I$ by $90^\circ$ ($X_C = 1/\omega C$).

$V_L$ and $V_C$ are in opposite directions, so the phasor sum gives

(b)(ii) Resonance

Resonance occurs in a series LCR circuit when the inductive reactance equals the capacitive reactance ($X_L = X_C$), so that the impedance is minimum ($Z = R$) and the current is maximum.

(b)(iii) Resonant frequency

(b)(iv) Quality factor

The quality factor measures the sharpness of resonance:

It is the ratio of the voltage across the inductor (or capacitor) at resonance to the applied voltage.

(b) $V/\sqrt{2}$

For a sinusoidal emf $V_0\sin\omega t$, the rms value is $V_{rms} = V_0/\sqrt2 \approx 0.707\,V_0$. It does not depend on the frequency of rotation, so with peak value $V$ the rms value is $V/\sqrt2$.

(a) In a series circuit the same current $I$ flows in all elements, so $I$ is the reference phasor, drawn along the horizontal axis. Then:

• $V_R = IR$ is drawn along $I$ (in phase with it).
• $V_L = IX_L$ is drawn $90^\circ$ ahead of $I$ (upward).
• $V_C = IX_C$ is drawn $90^\circ$ behind $I$ (downward).

The net reactive voltage is $V_L - V_C$ (taking $V_L>V_C$), vertical. The source voltage $V$ is the resultant of $V_R$ and $(V_L - V_C)$, making an angle $\phi$ with $I$, where $\tan\phi = \dfrac{X_L - X_C}{R}$.

(b) From the phasor diagram,

$V = \sqrt{V_R^2 + (V_L - V_C)^2} = I\sqrt{R^2 + (X_L - X_C)^2}$

$Z = \frac{V}{I} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$

(c) At resonance $X_L = X_C$:

$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.5\times20\times10^{-6}}} = \frac{1}{2\pi(3.16\times10^{-3})} \approx 50.3\ \text{Hz}$

(d) Tuning a radio or TV receiver: the LC circuit is adjusted so that its resonant frequency matches the frequency of the desired station, which then gives a maximum signal.

(e) A 220 V AC supply has a peak value of $220\sqrt2 \approx 311$ V, so it momentarily reaches a higher voltage than 220 V DC. Also, AC at 50 Hz reverses direction repeatedly and makes the muscles contract and relax repeatedly, which can disturb the heart rhythm and cause fibrillation. A person may be unable to let go of the conductor. DC causes a single jerk that tends to throw the person away from the conductor.

(B) pure resistive circuit

Average power is $P = V_{rms}I_{rms}\cos\phi$. A pure resistive circuit has $\phi = 0$, so $\cos\phi = 1$ and the power is maximum. A pure inductor or pure capacitor has $\cos\phi = 0$ (no power), and an LR circuit has $0 < \cos\phi < 1$.

(a) Choke coil

A choke coil is a coil of thick copper wire wound on a laminated soft-iron core. It has a large inductance but a very small resistance. It is used to control (reduce) alternating current in a circuit, for example in fluorescent tube lights, while wasting very little power.

(b) Why it is preferred over a resistor in an ac circuit

• If a resistor $R$ is used to reduce the current, it wastes energy as heat at the rate $I_{rms}^2R$.
• A choke reduces the current through its reactance $X_L = \omega L$. Because its resistance $r$ is very small, the phase angle is nearly $90^\circ$ and the power factor $\cos\phi = \dfrac{r}{\sqrt{r^2 + \omega^2L^2}} \approx 0$.
• So the average power lost in a choke, $P = V_{rms}I_{rms}\cos\phi$, is almost zero. A choke controls the current without wasting energy, which is why it is preferred.

(c) (i) Effect of increasing the frequency

   (a)               (b)                (c)
  --/\/\/--        --(((((--          ---| |---
  |   R   |        |   L   |          |   C   |
  --( ~ )--        --( ~ )--          --( ~ )--
      E                E                  E

At first all three currents are equal. When the frequency $f$ is increased:

• (a) Resistor: $R$ does not depend on frequency, so $I = \dfrac{E}{R}$ stays the same.
• (b) Inductor: $X_L = 2\pi fL$ increases with $f$, so $I = \dfrac{E}{2\pi fL}$ decreases. A faster change of current produces a larger back emf.
• (c) Capacitor: $X_C = \dfrac{1}{2\pi fC}$ decreases with $f$, so $I = E\,(2\pi fC)$ increases. The capacitor charges and discharges more times per second, so more charge flows.

(c) (ii) Replacing the ac emf by an equivalent dc emf

• Figure (a): A resistor offers the same opposition $R$ to both ac and dc. So the opposition, and the current, does not change.
• Figure (b): For dc, $f = 0$, so $X_L = 2\pi fL = 0$. No back emf is induced in steady dc, so the inductor offers only the small resistance of its wire (ideally zero). Its opposition falls greatly and the current becomes very large, much larger than in figure (a).

So for ac, both the resistor and the inductor oppose the current (here equally). For dc, only the resistor offers real opposition, and the inductor acts almost like a plain connecting wire.

Take $\varepsilon = \varepsilon_m\sin\omega t$ and $i = I\sin(\omega t - \phi)$, where $\phi$ is the phase constant, with $\tan\phi = \dfrac{X_L - X_C}{R}$.

(i) Sign of the phase constant

In the graph, the emf crosses zero (going up) and reaches its peak before the current does. So the current lags the emf, by roughly $80^\circ$. In $i = I\sin(\omega t - \phi)$, a lagging current means $\phi$ is positive. So the phase constant is positive, $\phi \approx +80^\circ$.

(ii) Phasor diagram

          ^  e_m  (emf phasor, longer)
          |
          |
          |   phi = about 80 deg
          |   (from I round to e_m)
          |              _.->  I  (current phasor, shorter)
          |        _.--''
          O__.--''-------------------->
            (both phasors rotate anticlockwise at w)

Both phasors start from the same origin O. The emf phasor is longer, and it is ahead of the current phasor by $\phi \approx 80^\circ$ in the anticlockwise direction of rotation.

(iii) More inductive or more capacitive?

The circuit is more inductive. Since $\phi > 0$, $\tan\phi = \dfrac{X_L - X_C}{R} > 0$, which gives $X_L > X_C$. In an inductive circuit the current lags the emf, as the graph shows.

(iv) Should $L$ be increased or decreased?

The average power delivered to the resistor is

This is largest at resonance, when $X_L = X_C$. Here $X_L > X_C$, so we must bring $X_L = \omega L$ down towards $X_C$. The inductance should be decreased. This reduces $(X_L - X_C)$ and $Z$, increases the current (and the power factor $\cos\phi$), and so increases the rate at which energy is transferred to the resistive load.