Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Dimension

RC → [T]

$\frac{L}{R}\to \left[T\right]$

$\sqrt[]{Lc}\to \left[T\right]$

$\frac{L}{C}\to$ dimensionless

Conservation of Energy b/w (1) & (2)

$\frac{1}{2}m{v}^2+mg\mathcal{l}=\frac{1}{2}m{u}^2$

$v=\sqrt[]{u^2-2g\mathcal{l}}$

$\overrightarrow{\Delta v}=v_{\widehat{j}}-u_{\widehat{i}}$

$\left|\Delta \overrightarrow{v}\right|=\sqrt[]{{\left(u\right)}^2+{\left(\sqrt[]{u^2-2g\mathcal{l}}\right)}^2}$

$=\sqrt[]{u^2+u^2-2g\mathcal{l}}$

$=\sqrt[]{2}\left(u^2-g\mathcal{l}\right)$

x = 2

The capacitive reactance is equal to the inductive reactance at resonance:

$X_L=X_C??=\frac{1}{\sqrt{L?C}}$

In an AC circuit, the power factor cos? measures the efficiency of power usage. Between the current and voltage, it is the cosine of the phase angle. A power factor of 1 means all the supplied power is used effectively and a low power factor means more energy is wasted.

Average power:

$P=V_{\text{rms}}\cdot I_{\text{rms}}\cdot cos??$

The alternating current's RMS (Root Mean Square) value or voltage offers the equivalent DC value that would generate the same power in a resistive circuit. The RMS is used in most practical measurements and is calculated as:

$I_{\text{rms}}=\frac{I_0}{\sqrt{2}},V_{\text{rms}}=\frac{V_0}{\sqrt{2}}$

Where ?0 and ?0 are the peak current and voltage, respectively.

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) For house hold supplies, AC currents are used which are having zero average value over a cycle.

The line is having some resistance so power factor cos φ = R/Z≠0

so, φ not equal to π /2 ⇒ φ < /2

i.e., phase lies between 0 and π /2.

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) When the AC voltage is applied to the capacitor, the plate connected to the positive terminal will be at higher potential and the plate connected to the negative terminal will be at lower potential.

The plate with positive charge will be at higher potential and the plate with negative charge will be at lower potential. So, we can say that the charge is in phase with the applied voltage.

P= E_rmsI_rmscos $\varnothing$

$\varnothing$ =90

So power, P = 0

This is a multiple choice answer as classified in NCERT Exemplar

According to the question power transferred Is P = I² Z cos?

as we know cos? = R/Z

R>0 and Z>0

cos? >0  so P>0

This is a multiple choice answer as classified in NCERT Exemplar

We have to transmit energy (power) over large distances at high alternating voltages, so current flowing through the wires will below because for a given power (P).

P= E_rmsI_rms, I_rms  is low when E_rms is high.

Power loss= I ² _rms R= low

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) According to the question, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreases as increasing frequency.

For a capacitive circuit,  capacitive reactance Xc= $\frac{1}{2\pi \vartheta \mathrm{C}}$

For an CR circuit Z= $\sqrt[]{R^2+\left(X_c^2\right)}$ when frequency increases X decreases.