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.

Check More: Class 12 Physics Notes

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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:

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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

1.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

2. Write any two features of nuclear forces.

3.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

4.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

6.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

2.
Write any two features of nuclear forces.

3.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

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