CBSE Class 12 Physics Chapter 7: Alternating Current - NCERT Solutions

Given values are:

Inductance, $L\; =\; 20\; \backslash text\{\; mH\}\; =\; 20\; \backslash times\; 10^\{-3\}\; \backslash text\{\; H\}$

Capacitance, $C\; =\; 50\; \backslash mu\; F\; =\; 50\; \backslash times\; 10^\{-6\}\; F$

Initial charge on the capacitor, $Q\; =\; 10\; \backslash text\{\; mC\}\; =\; 10\; \backslash times\; 10^\{-3\}\; \backslash text\{\; C\}$

(a) The total energy stored initially in the circuit is the energy stored in the capacitor when it has the maximum charge $Q$. This energy is given by the formula:

$$E\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{Q^2\}\{C\}$$

Substituting the given values:

$$E\; =\; \backslash frac\{\backslash left(10\; \backslash times\; 10^\{-3\}\backslash right)^2\}\{2\; \backslash times\; 50\; \backslash times\; 10^\{-6\}\}\; =\; \backslash frac\{100\; \backslash times\; 10^\{-6\}\}\{100\; \backslash times\; 10^\{-6\}\}\; =\; 1\; \backslash text\{\; J\}$$

Since the resistance of the circuit is negligible, there is no mechanism for energy loss. Therefore, the total energy stored in the LC circuit will be conserved during the oscillations. The energy will continuously transfer between the capacitor (as electrical energy) and the inductor (as magnetic energy), but the total amount of energy will remain constant at 1 J.

(b) The natural frequency of the LC circuit is given by the formula:

$v\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{LC\}\}$

Substituting the values of L and C:

$$v\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{(20\; \backslash times\; 10^\{-3\})\; \backslash times\; (50\; \backslash times\; 10^\{-6\})\}\}\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{1000\; \backslash times\; 10^\{-9\}\}\}\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{10^\{-6\}\}\}\; =\; \backslash frac\{1\}\{2\backslash pi\; \backslash times\; 10^\{-3\}\}\; =\; \backslash frac\{10^3\}\{2\backslash pi\}$$

$v\; \backslash approx\; \backslash frac\{1000\}\{6.283\}\; \backslash approx\; 159.15\; \backslash text\{\; Hz\}$

The natural angular frequency is:

$$\backslash omega\_r\; =\; \backslash frac\{1\}\{\backslash sqrt\{LC\}\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{(20\; \backslash times\; 10^\{-3\})\; \backslash times\; (50\; \backslash times\; 10^\{-6\})\}\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{10^\{-6\}\}\}\; =\; 10^3\; \backslash text\{\; rad/s\}$$

Thus, the natural frequency of the circuit is approximately 159.15 Hz, or the natural angular frequency is $10^3$ rad/s.

(c) The charge on the capacitor at any time $t$ in an LC circuit, starting with maximum charge $Q$ at $t=0$, is given by:

$$Q\text{'}(t)\; =\; Q\; \backslash cos(\backslash omega\_r\; t)$$

where $\backslash omega\_r\; =\; \backslash frac\{1\}\{\backslash sqrt\{LC\}\}\; =\; 10^3\; \backslash text\{\; rad/s\}$ is the angular frequency.

The time period of oscillation is $T\; =\; \backslash frac\{2\backslash pi\}\{\backslash omega\_r\}\; =\; \backslash frac\{2\backslash pi\}\{10^3\}\; \backslash approx\; 6.28\; \backslash text\{\; ms\}$.

(i) Energy stored completely electrical: This occurs when the energy in the capacitor is maximum, which happens when the charge $Q\text{'}(t)$ on the capacitor is maximum ($Q$ or $-Q$). This corresponds to $\backslash cos(\backslash omega\_r\; t)\; =\; \backslash pm\; 1$.

So, $\backslash omega\_r\; t\; =\; n\backslash pi$, where $n$ is an integer (0, 1, 2, ...).

Therefore, the times are $t\; =\; \backslash frac\{n\backslash pi\}\{\backslash omega\_r\}\; =\; \backslash frac\{n\; T\}\{2\}$. These times are $t\; =\; 0,\; \backslash frac\{T\}\{2\},\; T,\; \backslash frac\{3T\}\{2\},\; \backslash dots$

(ii) Energy stored completely magnetic: This occurs when the energy in the inductor is maximum, which happens when the charge $Q\text{'}(t)$ on the capacitor is zero ($Q\text{'}(t)\; =\; 0$). This corresponds to $\backslash cos(\backslash omega\_r\; t)\; =\; 0$.

So, $\backslash omega\_r\; t\; =\; \backslash frac\{(2n+1)\backslash pi\}\{2\}$, where $n$ is an integer (0, 1, 2, ...).

Therefore, the times are $t\; =\; \backslash frac\{(2n+1)\backslash pi\}\{2\backslash omega\_r\}\; =\; \backslash frac\{(2n+1)T\}\{4\}$. These times are $t\; =\; \backslash frac\{T\}\{4\},\; \backslash frac\{3T\}\{4\},\; \backslash frac\{5T\}\{4\},\; \backslash dots$

(d) The total energy $E$ is shared equally between the inductor and the capacitor when the electrical energy equals the magnetic energy. At this point, each form of energy is half of the total energy, i.e., $E/2$.

Electrical energy $E\_E\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{(Q\text{'}(t))^2\}\{C\}$ and Magnetic energy $E\_M\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{(I(t))^2\}\{L\}$.

We need $E\_E\; =\; E\_M\; =\; E/2$.

Since $E\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{Q^2\}\{C\}$, we need $E\_E\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{Q^2\}\{2C\}$. This means $\backslash frac\{1\}\{2\}\; \backslash frac\{(Q\text{'}(t))^2\}\{C\}\; =\; \backslash frac\{Q^2\}\{4C\}$, which simplifies to $(Q\text{'}(t))^2\; =\; \backslash frac\{Q^2\}\{2\}$, or $Q\text{'}(t)\; =\; \backslash pm\; \backslash frac\{Q\}\{\backslash sqrt\{2\}\}$.

Using $Q\text{'}(t)\; =\; Q\; \backslash cos(\backslash omega\_r\; t)$, we have $Q\; \backslash cos(\backslash omega\_r\; t)\; =\; \backslash pm\; \backslash frac\{Q\}\{\backslash sqrt\{2\}\}$, so $\backslash cos(\backslash omega\_r\; t)\; =\; \backslash pm\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}$.

This occurs when $\backslash omega\_r\; t\; =\; \backslash frac\{\backslash pi\}\{4\},\; \backslash frac\{3\backslash pi\}\{4\},\; \backslash frac\{5\backslash pi\}\{4\},\; \backslash frac\{7\backslash pi\}\{4\},\; \backslash dots$

In terms of the time period $T$, these times are $t\; =\; \backslash frac\{T\}\{8\},\; \backslash frac\{3T\}\{8\},\; \backslash frac\{5T\}\{8\},\; \backslash frac\{7T\}\{8\},\; \backslash dots$

(e) If a resistor is inserted in the circuit, the LC circuit becomes an RLC circuit. In such a circuit, the oscillations are damped due to the presence of resistance. The energy that was initially stored in the capacitor and inductor is gradually dissipated as heat in the resistor due to the flow of current. Over a long period, all the initial energy stored in the circuit will be dissipated as heat, and the oscillations will eventually die out. The total energy eventually dissipated as heat will be equal to the initial total energy stored in the circuit, which is 1 Joule.