CBSE Class 12 Physics Chapter 7 Alternating Current NCERT Solutions

Given the frequency of the AC circuit,

u = 50 \text{ Hz}, and the RMS current, I_{\text{rms}} = 5 \text{ A}. We need to find the instantaneous current at time t = \frac{1}{300} \text{ s} after the current passes through zero.

First, calculate the peak current (I_0) from the RMS current:

I_{\text{rms}} = \frac{I_0}{\sqrt{2}}

I_0 = I_{\text{rms}} \times \sqrt{2} = 5\sqrt{2} \text{ A}

The angular frequency (\omega) is given by:

\omega = 2\pi

u = 2\pi \times 50 = 100\pi \text{ rad/s}

The instantaneous current (I(t)) in an AC circuit can be represented as I(t) = I_0 \sin(\omega t), assuming the current is zero at t=0 and increasing.

Substitute the values at t = \frac{1}{300} \text{ s}:

I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(100\pi \times \frac{1}{300}\right)

I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(\frac{\pi}{3}\right)

Since \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}:

I\left(\frac{1}{300}\right) = 5\sqrt{2} \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{6}}{2} = 5\sqrt{\frac{6}{4}} = 5\sqrt{\frac{3}{2}} \text{ A}

Therefore, the value of the current at t = \frac{1}{300} \text{ s} is 5\sqrt{3/2} \text{ A}.

For maximum power to be delivered from the generator to the load, the impedance of the load must be the complex conjugate of the impedance of the source (generator plus internal components).

The impedance of the generator is $Z\_g\; =\; R\_g\; +\; iX\_g$.

The impedance of the load is $Z\_L\; =\; R\_L\; +\; iX\_L$.

For maximum power transfer, the condition is that the real parts of the impedances should be equal, and the imaginary parts should be opposite:

$R\_L\; =\; R\_g$

$X\_L\; =\; -X\_g$

This condition ensures that the total reactance in the circuit is zero, and the power transfer is maximized when the resistance of the load matches the internal resistance of the generator.

Therefore, for maximum power transfer, the value of $X\_L$ must be equal to $-X\_g$.

AC voltmeters and ammeters are designed to measure the RMS (Root Mean Square) value of the alternating voltage or current. The RMS value is the effective value of the AC, which produces the same amount of heat as an equivalent DC value.

The RMS value is calculated as the square root of the mean of the square of the instantaneous voltage over one cycle. Mathematically, for a voltage $v(t)$, the RMS voltage is $V\_\{\backslash text\{rms\}\}\; =\; \backslash sqrt\{\backslash langle\; v(t)^2\; \backslash rangle\}$.

Therefore, when the meter shows a steady input voltage of 220 V, it means the RMS value of the AC mains voltage is 220 V. The meter is calibrated to display this RMS value, which is derived from the mean of the squared instantaneous voltages.

The resonant frequency (

u_0) of an L-C-R series circuit is given by the formula:

u_0 = \frac{1}{2\pi\sqrt{LC}}

To reduce the resonant frequency (

u_0), we need to either increase the inductance (L) or increase the capacitance (C).

Let's analyze the options:

(a) Reducing the generator frequency does not change the resonant frequency of the circuit itself; it only affects whether the circuit is at resonance or not.

(b) Adding another capacitor in parallel to the existing capacitor increases the total capacitance (C_{\text{total}} = C_1 + C_2). An increase in C will decrease

u_0 according to the formula.

(c) Removing the iron core of the inductor would decrease the inductance (L), which would increase the resonant frequency.

(d) Removing the dielectric from the capacitor would decrease its capacitance (C), which would increase the resonant frequency.

Therefore, the correct way to reduce the resonant frequency is to add another capacitor in parallel.

For better tuning of an L-C-R circuit used for communication, the quality factor ($Q$) of the circuit must be as high as possible. A high quality factor means the circuit is more selective and can distinguish between closely spaced frequencies.

The quality factor ($Q$) for a series L-C-R circuit is given by the formula:

$Q\; =\; \backslash frac\{1\}\{R\}\; \backslash sqrt\{\backslash frac\{L\}\{C\}\}$

To maximize $Q$, we need to:

• Minimize the resistance ($R$).
• Maximize the inductance ($L$).
• Minimize the capacitance ($C$).

Let's evaluate the $Q$ factor for each option:

(a) $Q\; =\; \backslash frac\{1\}\{20\}\; \backslash sqrt\{\backslash frac\{1.5\}\{35\; \backslash times\; 10^\{-6\}\}\}\; \backslash approx\; 0.05\; \backslash times\; \backslash sqrt\{42857\}\; \backslash approx\; 0.05\; \backslash times\; 207\; \backslash approx\; 10.35$

(b) $Q\; =\; \backslash frac\{1\}\{25\}\; \backslash sqrt\{\backslash frac\{2.5\}\{45\; \backslash times\; 10^\{-6\}\}\}\; \backslash approx\; 0.04\; \backslash times\; \backslash sqrt\{55555\}\; \backslash approx\; 0.04\; \backslash times\; 235.7\; \backslash approx\; 9.43$

(c) $Q\; =\; \backslash frac\{1\}\{15\}\; \backslash sqrt\{\backslash frac\{3.5\}\{30\; \backslash times\; 10^\{-6\}\}\}\; \backslash approx\; 0.0667\; \backslash times\; \backslash sqrt\{116667\}\; \backslash approx\; 0.0667\; \backslash times\; 341.6\; \backslash approx\; 22.79$

(d) $Q\; =\; \backslash frac\{1\}\{25\}\; \backslash sqrt\{\backslash frac\{1.5\}\{45\; \backslash times\; 10^\{-6\}\}\}\; \backslash approx\; 0.04\; \backslash times\; \backslash sqrt\{33333\}\; \backslash approx\; 0.04\; \backslash times\; 182.6\; \backslash approx\; 7.30$

Comparing the calculated quality factors, option (c) yields the highest value of $Q$ (approximately 22.79). This combination has the lowest resistance and a good balance of inductance and capacitance for a high quality factor.

Therefore, the combination in option (c) should be selected for better tuning.