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, V_R= i_R … [1]

$\therefore$ Because voltage across resistance and current always are in phase, thereby the vectors of the voltage of resistance (V_R) 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, X_C= 1/ω_C … [2]

Here,

ω = Angular Frequency of the Source
C = Capacitance

Therefore, magnitude of the given voltage vector of capacitance equals, V_C= iX_C… [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, V_net = V_R+ V_C.

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

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

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

⇒ V_net = √(iR)²+ (iXC)²

⇒ V_net=i√R²+ 1/ω²C²

Here, V_net= 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.

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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:

Related Questions

CBSE CLASS XII Related Questions

3.A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

4.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 ?

5.With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

3.
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

4.
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 ?

5.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.