CBSE Class 12 Physics Chapter 14: Alternating Current NCERT Solutions

Given: Resistance, $R\; =\; 100\; \backslash Omega$ Supply voltage (rms), $V\; =\; 220\; \backslash text\{\; V\}$ Frequency, $v\; =\; 50\; \backslash text\{\; Hz\}$

1. The rms value of current ($I$) in a purely resistive circuit is calculated using Ohm's law: $I\; =\; \backslash frac\{V\}\{R\}$. Substituting the given values: $I\; =\; \backslash frac\{220\; \backslash text\{\; V\}\}\{100\; \backslash Omega\}\; =\; 2.2\; \backslash text\{\; A\}$ Therefore, the rms value of the current in the circuit is 2.2 A.
2. The net power consumed over a full cycle in a purely resistive AC circuit is given by the formula $P\; =\; V\; \backslash times\; I$, where V and I are the rms values. Substituting the values: $P\; =\; 220\; \backslash text\{\; V\}\; \backslash times\; 2.2\; \backslash text\{\; A\}\; =\; 484\; \backslash text\{\; W\}$ Thus, the net power consumed over a full cycle is 484 W.

1. Given the peak voltage, $V\_0\; =\; 300\; \backslash text\{\; V\}$. The relationship between peak voltage ($V\_0$) and rms voltage ($V$) is $V\; =\; \backslash frac\{V\_0\}\{\backslash sqrt\{2\}\}$. Calculating the rms voltage: $V\; =\; \backslash frac\{300\; \backslash text\{\; V\}\}\{\backslash sqrt\{2\}\}\; \backslash approx\; \backslash frac\{300\}\{1.414\}\; \backslash approx\; 212.1\; \backslash text\{\; V\}$ So, the rms voltage is approximately 212.1 V.
2. Given the rms value of current, $I\; =\; 10\; \backslash text\{\; A\}$. The relationship between peak current ($I\_0$) and rms current ($I$) is $I\_0\; =\; \backslash sqrt\{2\}\; \backslash times\; I$. Calculating the peak current: $I\_0\; =\; \backslash sqrt\{2\}\; \backslash times\; 10\; \backslash text\{\; A\}\; \backslash approx\; 1.414\; \backslash times\; 10\; \backslash text\{\; A\}\; =\; 14.14\; \backslash text\{\; A\}$ Therefore, the peak current is approximately 14.14 A.

Given:

Inductance, L = 44 \text{ mH} = 44 \times 10^{-3} \text{ H}

Supply voltage (rms), V = 220 \text{ V}

Frequency, v = 50 \text{ Hz}

First, calculate the angular frequency (\omega):

\omega = 2\pi v = 2\pi \times 50 \text{ Hz} = 100\pi \text{ rad/s}

Next, calculate the inductive reactance (X_L), which is the opposition offered by the inductor to the AC current:

X_L = \omega L = (100\pi \text{ rad/s}) \times (44 \times 10^{-3} \text{ H})

X_L = 4.4\pi \Omega \approx 4.4 \times 3.14159 \Omega \approx 13.82 \Omega

Finally, calculate the rms value of the current (I) using Ohm's law for AC circuits:

I = \frac{V}{X_L} = \frac{220 \text{ V}}{13.82 \Omega} \approx 15.92 \text{ A}

Hence, the rms value of the current in the circuit is approximately 15.92 A.

Given:

Capacitance, C = 60 \mu F = 60 \times 10^{-6} F

Supply voltage (rms), V = 110 \text{ V}

Frequency, v = 60 \text{ Hz}

First, calculate the angular frequency (\omega):

\omega = 2\pi v = 2\pi \times 60 \text{ Hz} = 120\pi \text{ rad/s}

Next, calculate the capacitive reactance (X_C), which is the opposition offered by the capacitor to the AC current:

X_C = \frac{1}{\omega C} = \frac{1}{(120\pi \text{ rad/s}) \times (60 \times 10^{-6} \text{ F})}

X_C = \frac{1}{7200\pi \times 10^{-6}} \Omega = \frac{10^6}{7200\pi} \Omega \approx \frac{1000000}{22619.47} \Omega \approx 44.21 \Omega

Finally, calculate the rms value of the current (I) using Ohm's law for AC circuits:

I = \frac{V}{X_C} = \frac{110 \text{ V}}{44.21 \Omega} \approx 2.49 \text{ A}

Therefore, the rms value of the current in the circuit is approximately 2.49 A.