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**3. SERIES AC CIRCUIT**

**3.1 When only Resistance is in an AC circuit**

Consider a simple ac circuit consisting of a resistor of resistance R and an ac generator, as shown in the figure. According to Kirchhoff's loop law at any instant, the algebraic sum of the potential difference around a closed loop in a circuit must be zero.

ε - V_{R} = 0

ε - i_{R}R = 0

ε_{0} sinωt - i_{R}R = 0

i_{R} = sinωt = *i*_{0} sin ωt ..(i)

where *i*_{0} is the maximum current. *i*_{0} =

From above equations, we see that the instantaneous voltage drop across the resistor

V_{R} = i_{0}R sinωt ...(ii)

We see in equation (i) & (ii), i_{R} and V_{R} both vary as sin wt and reach their maximum values at the same time as shown in figure (a), they are said to be in phase. A phasor diagram is used to represent phase relationships. The lengths of the arrows correspond to V_{0} and i_{0}. The projections of the arrows onto the vertical axis give V_{R} and i_{R}. In case of the single-loop resistive circuit, the current and voltage phasors lie along the same line, as shown in figure (b), because *i*_{R} and *V*_{R} are in phase.

**3.2 When only Inductor is in An AC circuit**

Now consider an ac circuit consisting only of an Inductor of inductance L connected to the terminals of an ac generator, as shown in the figure. The induced emf across the inductor is given by Ldi/dt. On applying Kirchhoff's loop rule to the circuit

ε- V_{L} = 0 ⇒ ε - L

When we rearrange this equation and substitute

ε = ε_{0} sin ωt, we get

= ε_{0} sin ωt ...(iii)

Integration of this expression gives the current as a function of time

*i*_{L} = = -

For average value of current over one time period to be zero, C = 0

Therefore, *i*_{L} = -

When we use the trigonometric identity

coswt = - sin(wt - p/2), we can express equation as

*i*_{L} = ...(iv)

From equation (iv), we see that the current reaches its maximum values when cos wt = 1.

*i*_{0} = = ...(v)

where the quantity X_{L}, called the inductive reactance, is

X_{L} = ωL

The expression for the rms current is similar to equation (v), with ε_{0} replaced by ε_{rms}.

Inductive reactance, like resistance, has unit of ohm.

We can think of equation (v) as Ohm's law for an inductive circuit.

On comparing result of equation (iv) with equation (iii), we can see that the current and voltage are out of phase with each other by π/2 rad, or 90º. A plot of voltage and current versus time is given in figure (a). The voltage reaches its maximum value one quarter of an oscillation period before the current reaches its maximum value. The corresponding phasor diagram for this circuit is shown in figure (b). Thus, we see that for a sinusoidal applied voltage, the current in an inductor always lags behind the voltage across the inductor by 90º.

*Ex.4 An inductor of inductance L = 5 H is connected to an*

*AC source having voltage v = 10 sin (10t + )*

*Find *

*(i) Inductive Reactance (x _{L})*

*(ii) Peak & Rms voltage (V _{0} & V_{rms})*

*(iii) Peak & Rms current (I _{0} & I_{rms})*

*(iv) Instantaneous current (I _{(t)})*

**Sol.** (i) x_{L} = ωL_{ }= 10 × 5 = 50

(ii) v_{0} = 10

v_{rms} =

(iii)

I_{rms} =

(iv) I(t) =

**3.3 When only Capacitor is in An AC circuit**

Figure shows an ac circuit consisting of a capacitor of capacitance C connected across the terminals of an ac generator. On applying Kirchhoff's loop rule to this circuit, we get

ε - V_{C} = 0

V_{C} = ε = ε_{0} sin ωt ...(vi)

where V_{C} is the instantaneous voltage drop across the capacitor. From the definition of capacitance, V_{C} = Q/C, and this value for V_{C} substituted into equation gives

Q = C ε_{0} sin ωt

Since i = dQ/dt, on differentiating above equation gives the instantaneous current in the circuit.

Here again we see that the current is not in phase with the voltage drop across the capacitor, given by equation (vi). Using the trigonometric identity cos ωt = sin(ωt + π/2), we can express this equation in the alternative from

...(vii)

From equation (vii), we see that the current in the circuit reaches its maximum value when cos ωt= 1.

Where X_{C} is called the capacitive reactance.

The SI unit of X_{C} is also ohm. The rms current is given by an expression similar to equation with V_{0} replaced by V_{rms}.

Combining equation (vi) & (vii), we can express the instantaneous voltage drop across the capacitor as

V_{C} = V_{0} sin ωt = i_{0} X_{C} sin ωt

Comparing the result of equation (v) with equation (vi), we see that the current is π/2 rad = 90º out of phase with the voltage across the capacitor. A plot of current and voltage versus time, shows that the current reaches its maximum value one quarter of a cycle sooner than the voltage reaches its maximum value. The corresponding phasor diagram is shown in the figure (b). Thus we see that for a sinusoidally applied emf, the current always leads the voltage across a capacitor by 90º.

**Brain Teaser**

*What is the reactance of a capacitor connected to a constant DC source ?*

**Ex.5 A capacitor of capacitive reactance 5? is connected with A.C. source having voltage V = 3 sin (ωt + p/6). Find rms and Peak voltage rms and peak current and instantaneous current**

**Sol. **On comparing with

⇒ v_{0} = 3

⇒

⇒ I(t) = I_{0} sin

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