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Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET PDF Download

Table of contents
Series L-R Circuit
Series C-R Circuit
L.C. Circuit
Series L-C-R Circuit

Series L-R Circuit

Now consider an ac circuit consisting of a resistor of resistance R and an inductor of inductance L in series with an ac source generator.
Suppose in phasor diagram, current is taken along positive x-direction. The V_R is also along positive x-direction and V_L along positive y-direction as we know that potential difference across a resistance in ac is in phase with current and it leads in phase by 90º with current across the inductor, and as we know V_R = i₀R & V₀ = i₀X_L
i = i₀ sin wt

V_R(t) = i₀ Rsin ωt
V_L(t) = i₀ X_L sin (ωt + p/2)
hence we can write
V(t) = i₀R sin ωt + i₀ X_L sin (ωt + p/2)
V₀ = i₀
where is known as impedence (z) of the circuit.
now we can write

where tan β=
hence β =

Example 1. When 100 volt dc is applied across a coil, a current of 1 amp flows through it; when 100 V ac of 50 Hz is applied to the same coil, only 0.5 amp flows. Calculate the resistance of inductance of the coil.

Sol. In case of a coil, i.e., L - R circuit.

i = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET with Z = =

So when dc is applied, ω = 0, so z = R

and hence i = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET i.e., R = = = 200 ?

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET i.e., ω²L² = Z²-R²

i.e., (2πfL)² = 200² - 100² = 3 × 10⁴ (as ω = 2πf)

So, L = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = = 0.55 H

Example 2. A 12 ohm resistance and an inductance of 0.05/p henry with negligible resistance are connected in series. Across the end of this circuit is connected a 130 volt alternating voltage of frequency 50 cycles/second. Calculate the alternating current in the circuit and potential difference across the resistance and that across the inductance.

Sol. The impedance of the circuit is given by

Z = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET =

= Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = = 13 ohm

Current in the circuit i = E/Z = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 10 amp

Potential difference across resistance

V_R = iR = 10 × 12 = 120 volt

Inductive reactance of coil X_L = ωL = 2πfL

Therefore, X_L = 2π × 50 × Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 5 ohm

Potential difference across inductance

V_L = i × X_L = 10 × 5 = 50 volt

Series C-R Circuit

Now consider an ac circuit consisting of a resistor of resistance R and an capacitor of capacitance C in series with an ac source generator.

Suppose in phasor diagram current is taken along positive x-direction. Then V_R is also along positive x-direction but V_C is along negative y-direction as potential difference across a capacitor in ac lags in phase by 90º with the current in the circuit. So we can write,

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

V_R = I₀ R sin ωt Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Potential difference across capacitor

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Potential at any instant t

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

V(t) = V₀ sin (ωt + b)

tan α = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Example 3. An A.C. source of angular frequency w is fed across a resistor R and a capacitor C in series. The current registered is i. If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency.

Sol. At angular frequency w, the current in R-C circuit is given by

i_rms = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET ...(i)

When frequency is changed to ω/3, the current is halved. Thus

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = ...(ii)

From equations (i) and (ii), we have

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET =

Solving this equation, we get Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Hence, the ratio of reactance to resistance is Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Example 4. A 50 W, 100 V lamp is to be connected to an ac mains of 200 V, 50 Hz. What capacitance is essential to be put in series with the lamp?

Sol. As resistance of the lamp R = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 200 ? and the maximum current i = = = ; so when the lamp is put in series with a capacitance and run at 200 V ac, from V =iZ we have,

Z = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = = 400?

Now as in case of C-R circuit, Z = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET ,

i.e., R² + Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 160000

or, Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 16 × 10⁴ - (200)² = 12 × 10⁴

So, Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = × 10²

or C = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

i.e., C = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 9.2 μF

L.C. Circuit

As shown in figure a capacitor and inductance are connected in series method and alternating voltage is applied across the circuit.

Let X_c is capacitance reactance,

X_L is Inductance reactance,

i = i₀ sin wt current flowing through the circuit

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

V_C(t) = i₀ X_C sin (ωt - π/2)

V_L(t) = i₀ X_L sin (ωt + π/2)

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

= i₀ X_C sin wt cos π/2 - i₀ X_C cos wt sin π/2 + i₀ X_L sin wt cos π/2 + i₀ X_L cos wt sin π/2

= i₀ cos w t(X_L - X_C)

V(t) = V₀sin (wt + p/2)

V₀ = i₀Z Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Z = (X_L- X_C)

cos Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET = 0

V_CO = i₀ X_C ; Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Series L-C-R Circuit

Now consider an ac circuit consisting of a resistor of resistance R, a capacitor of capacitance C and an inductor of inductance L are in series with an ac source generator.

Suppose in a phasor diagram current is taken along positive x-direction. Then V_R is along positive x-direction, V_L along positive y-direction and V_C along negative y-direction, as potential difference across an inductor leads the current by 90° in phase while that across a capacitor, lags by 90°.

V = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

L - R - C circuit

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Impedance phasor of above circuit

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

& Impedance triangle

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

here B is phase angle By triangle tan β= Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Power factor cos Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET =

Let I be the current in the series circuit of any instant then

(1) Voltage V(t) = V₀ sin (ωt + β) = i₀ z sin (ωt + β)

here v₀ = i₀z & v_rms = i_rmsz

(2) Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET here voltage V_L across the inductance is ahead of current I in phase by π/2 rad

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

(3) V_C(t) = V_{O C} sin (ωt - π/2)

here voltage V_C across the capacitance lags behind the current I in phase by π/2 rad

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

(4) V_R(t) = i₀ R sin ωt

here voltage V_R across the resistor R has same phase as I

V_{O R} = I_OR

Special Case :

(1) When X_L > X_C or V_L > V_C then emf is ahead of current by phase β which is given by

tan β = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET or cos f =

The series LCR circuit is said to be inductive

(2) When X_L < X_C or V_L < V_C then current is ahead of emf by phase angle β which is given by

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET or

The series LCR circuit is said to be capacitive

(3) When X_L = X_C or V_L = V_C, b = 0, the emf and current will be in the same phase. The series LCR circuit is said to be purely resistive. It may also be noted that

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET or or I_Rms =

Susceptance : The reciprocal of the reactane of an a.c. circuit is called its susceptance.

Admittance : The reciprocal of the impedance of an a.c. circuit is called its admittance.

Ex.10 Figure shows a series LCR circuit connected to a variable voltage source V = 10 sin (ωt + p/4) ;

x_L = 10 ?, X_C = 6 ?, R = 3 ?

Calculate Z, i₀, i_rms, v_rms, V_{L O}, V_{C O}, V_{R O},b,

V_{L Rms}, V_{C Rms,}V_Rms, i(t), V_L(t), V_c(t),and V_R(t)

X_L > X_C

Sol. V = 10 sin (wt + p/4) so V₀ = 10 volt

V_rms = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Therefore, Z = Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET ;

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

i(t) = 20 sin (ωt + π/4 - 53°)

V_L(t) = 20 sin (ωt + π/4 - 53° + π/2)

= 20 sin (ωt + Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET - 53° - ) = 12 sin (ωt - - 53°)

Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET

Ex.11 A resistor of resistance R, an inductor of inductance L and a capacitor of capacitance C all are connected in series with an a.c. supply. The resistance of R is 16 ohm and for a given frequency the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find

(a) the potential difference across R, L and C (b) the impedance of the circuit

(c) the voltage of a.c. supply (d) phase angle

Sol. (a) Potential difference across resistance

V_R = iR = 5 × 16 = 80 volt

Potential difference across inductance

V_L = i × (wL) = 5 × 24 = 120 volt

The document Series L-R, C-R & C-R Circuit | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.

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FAQs on Series L-R, C-R & C-R Circuit - Physics Class 12 - NEET

1. What is a Series L-R circuit?

Ans. A Series L-R circuit is a type of electrical circuit that consists of an inductor (L) and a resistor (R) connected in series. In this circuit, the inductor opposes changes in current flow due to its property of storing energy in its magnetic field, while the resistor dissipates electrical energy in the form of heat.

2. What is the significance of a Series C-R circuit?

Ans. A Series C-R circuit is a circuit that includes a capacitor (C) and a resistor (R) connected in series. This type of circuit is commonly used in power factor correction applications. The capacitor in the circuit helps to improve the power factor by compensating for the reactive power in the system.

3. What is an L.C. circuit?

Ans. An L.C. circuit, also known as a tank circuit or resonant circuit, is an electrical circuit that includes an inductor (L) and a capacitor (C) connected in parallel. This type of circuit exhibits resonance, which occurs when the natural frequency of the circuit matches the frequency of an applied alternating current. L.C. circuits are commonly used in radio receivers and transmitters.

4. What is the purpose of a Series L-C-R circuit?

Ans. A Series L-C-R circuit is designed to provide a specific frequency response or filter a particular range of frequencies. By combining the elements of an inductor (L), a capacitor (C), and a resistor (R) in series, this circuit can be used to create low-pass, high-pass, or band-pass filters. The values of the components determine the cutoff frequency and the shape of the frequency response.

5. What are some applications of Series L-R, C-R, and L-C-R circuits?

Ans. Series L-R, C-R, and L-C-R circuits find applications in various areas of electrical engineering. Some common applications include audio frequency amplifiers, radio frequency filters, power factor correction in electrical distribution systems, impedance matching, and resonance control in electronic devices. These circuits are also used in communication systems, signal processing, and power electronics.

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