A fixed amplitude V0 and AC voltage source of variable angular frequency are joined in series with capacitance C and an electric bulb of resistance R (inductance zero). When variable angular frequency is increased:

As per the given values, When Variable Angular Frequency ω is increased, the following can occur:

The amplitudes of the voltages, in an AC circuit, across the components and the current are treated as vectors. Simply, the phase difference between voltage and current is apparently the angle between the vectors.

Now, assume the amplitude of the flowing current in the circuit = i

Therefore, magnitude of voltage vector of given resistance, VR = iR … [1]

\(\therefore\) Because voltage across resistance and current always are in phase, thereby the vectors of the voltage of resistance (VR) and current (i) are found to be parallel.

Capacitance, in an AC circuit, serves as resistance to the circuit’s flow, also called the reactance of the circuit (Xc).

Thus,

Value of capacitive reactance equals, XC = 1/ωC … [2]

Here,

  • ω = Angular Frequency of the Source
  • C = Capacitance

Therefore, magnitude of the given voltage vector of capacitance equals, VC = iXC … [3]

A phase difference of −π/2 between the capacitor voltage and the current is found. The capacitor voltage lags the current behind by a phase of π/2. Thus, the vector of capacitor voltage (Vc) is found at an angle of −π/2 from the current vector.

Therefore, the resultant or the net voltage in the circuit is equal to, Vnet = VR + VC.

Note: “VR and VC are vectors and hence, Vnet is also a vector.”

Therefore, the magnitude of net voltage equals, Vnet = √V2R + V2C 

Now, after value substitution of VR and VC from equations [1] and [3].

⇒ Vnet = √(iR)2 + (iXC)2

⇒ Vnet=i√R2 + 1/ω2C2

Here, Vnet = V0

⇒ I = V0/√R2 + 1ω2C2

Considering the value of ω is increased, the value of the current [1] will also increase. Now, the dissipated power via the resistance is expressed as P = i2R. Thus, the power in the given circuit will also rise. Now, if the power of the bulbs increases, it is more likely to glow brighter.

Check More: Class 12 Physics Notes


Related Questions

  1. In A Graph Between Current I And Voltage V, Find The Portion Corresponding To Negative Resistance.
  2. A Closed Coil Has 500 Turns Across Rectangular Frame Of Area 4.0 Cm2 With Resistance Of 500 Ohms. The Coil Is Plane Perpendicular To A Uniform Magnetic Field Of 0.2wb/M2. Find Amount Of Charge Through Coil If Turned Over (180 Degrees Rotation).
  3. 1.0 M Rectangular Loop With A Sliding Connector Is In Uniform Magnetic Field 2t Perpendicular To Plane Of Loop. Resistance Is 2 Ohms. Two Resistances, 6 Ohms And 3 Ohms, Are Connected.
  4. Resistance In Meter Bridge’s Two Arms Are 5 Ohms And R Ohms. When Resistance R Is Shunted With Equal Resistance, New Balance Point Becomes 1.6l1. Calculate R.
  5. If R, C And L Are Fundamental Quantities In A Circuit Like Resistance, Capacitance And Inductance In W, Then Find Dimensional Formula For Resistance And Capacitance.
  6. Two Copper Wires, Of 1 M And 9 M Respectively, Have Same Resistance. Find Their Diameters In Ratio.
  7. Two Identical Resistors With Resistances 15 Ohm Are Connected In Series And Parallel To A Battery Of 6 V. Calculate Ratio Of Power Consumed.
  8. Two Concentric Coplanar Circular Loops Of Wire (Resistance Per Unit Length 10 4 Ohms M-1) Have Diameters 0.2 M And 2 M. With Time-Varying Potential Difference Of (4+2.5t) Applied To Larger Loop, Find Current In Smaller One.
  9. Three Incandescent Bulbs (Each 100 W) Are Attached In Series. In Another Circuit, Three More Bulbs Of Same Wattage Are Attached Parallelly To An Equal Source.
  10. A Circuit Consists Of A Battery Of 3 Cells (2 V Each), A Combination Of Three Resistors, 10 Ohm, 20 Ohm And 30 Ohm, Attached Parallelly, With Plug Key And Ammeter (In Series).
  11. For The Resistor Combination, Find The Equivalent Resistance Between M and N.

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CBSE CLASS XII Related Questions

  • 1.
    Read the following paragraph and answer the questions that follow.
    A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.


      • 2.
        Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.


          • 3.
            Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

              • \( (M_x - M_z)<M_y \)
              • \( (M_x - M_y)<M_z \)
              • \( M_x>(M_y + M_z) \)
              • \( M_x<(M_y + M_z) \)

            • 4.
              Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.


                • 5.
                  Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

                    • The total charge of the two spheres is conserved.
                    • Both spheres attain the same potential.
                    • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
                    • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$

                  • 6.
                    Write two advantages of reflecting telescope over refracting telescope.

                      CBSE CLASS XII Previous Year Papers

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