CBSE Class 12 Physics Chapter 13: Alternating Current NCERT Solutions

Given: 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, $Q\; =\; 10\; \backslash text\{\; mC\}\; =\; 10\; \backslash times\; 10^\{-3\}\; \backslash text\{\; C\}$

1. The total energy stored initially in the LC circuit is the energy stored in the capacitor when it has the maximum charge. This is calculated using the formula for the energy stored in a capacitor: $$E\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{Q^2\}\{C\}$$ Substituting the given values: $$E\; =\; \backslash frac\{\backslash left(10\; \backslash times\; 10^\{-3\}\backslash right)^2\}\{2\; \backslash times\; 50\; \backslash times\; 10^\{-6\}\}\; =\; \backslash frac\{100\; \backslash times\; 10^\{-6\}\}\{100\; \backslash times\; 10^\{-6\}\}\; =\; 1\; \backslash text\{\; J\}$$ Thus, the total energy stored initially is 1 Joule. Since the resistance of the circuit is negligible, there is no mechanism for energy loss. The energy oscillates between being stored as electric field energy in the capacitor and magnetic field energy in the inductor. Therefore, the total energy stored is conserved during LC oscillations.

(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: $$v\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{(20\; \backslash times\; 10^\{-3\})\; \backslash times\; (50\; \backslash times\; 10^\{-6\})\}\}\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{1000\; \backslash times\; 10^\{-9\}\}\}\; =\; \backslash frac\{1\}\{2\backslash pi\backslash sqrt\{10^\{-6\}\}\}\; =\; \backslash frac\{1\}\{2\backslash pi\; \backslash times\; 10^\{-3\}\}\; =\; \backslash frac\{1000\}\{2\backslash pi\}\; \backslash text\{\; Hz\}$$ Calculating the approximate value: $v\; \backslash approx\; \backslash frac\{1000\}\{6.283\}\; \backslash approx\; 159.2\; \backslash text\{\; Hz\}$ The natural angular frequency is given by: $\backslash omega\_r\; =\; \backslash frac\{1\}\{\backslash sqrt\{LC\}\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{(20\; \backslash times\; 10^\{-3\})\; \backslash times\; (50\; \backslash times\; 10^\{-6\})\}\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{10^\{-6\}\}\}\; =\; 10^3\; \backslash text\{\; rad/s\}$ So, the natural frequency of the circuit is approximately 159.2 Hz, and the natural angular frequency is $1000\; \backslash text\{\; rad/s\}$.

1. The charge $Q\text{'}(t)$ on the capacitor at any time $t$ in an LC circuit with initial charge $Q$ is given by $Q\text{'}(t)\; =\; Q\; \backslash cos(\backslash omega\_r\; t)$, where $\backslash omega\_r\; =\; \backslash frac\{1\}\{\backslash sqrt\{LC\}\}$ is the angular frequency. The time period of oscillation is $T\; =\; \backslash frac\{2\backslash pi\}\{\backslash omega\_r\}\; =\; 2\backslash pi\backslash sqrt\{LC\}$. (i) The energy stored is completely electrical when it is entirely in the capacitor. This occurs when the charge on the capacitor is maximum, i.e., $Q\text{'}(t)\; =\; Q$. This happens when $\backslash cos(\backslash omega\_r\; t)\; =\; 1$, which is true for $\backslash omega\_r\; t\; =\; 0,\; 2\backslash pi,\; 4\backslash pi,\; \backslash dots$. Therefore, the times are $t\; =\; 0,\; \backslash frac\{T\}\{2\},\; T,\; \backslash frac\{3T\}\{2\},\; \backslash dots$. (ii) The energy stored is completely magnetic when it is entirely in the inductor. This occurs when the charge on the capacitor is zero, i.e., $Q\text{'}(t)\; =\; 0$. This happens when $\backslash cos(\backslash omega\_r\; t)\; =\; 0$, which is true for $\backslash omega\_r\; t\; =\; \backslash frac\{\backslash pi\}\{2\},\; \backslash frac\{3\backslash pi\}\{2\},\; \backslash frac\{5\backslash pi\}\{2\},\; \backslash dots$. Therefore, the times are $t\; =\; \backslash frac\{T\}\{4\},\; \backslash frac\{3T\}\{4\},\; \backslash frac\{5T\}\{4\},\; \backslash dots$.

1. The total energy $E$ is equally shared between the inductor and the capacitor when the electrical energy equals the magnetic energy. At this point, the charge on the capacitor $Q\text{'}(t)$ is such that the energy stored in the capacitor $E\_C\; =\; \backslash frac\{1\}\{2\}\; \backslash frac\{(Q\text{'}(t))^2\}\{C\}$ is half of the total initial energy $E$. So, $E\_C\; =\; \backslash frac\{E\}\{2\}$. $\backslash frac\{1\}\{2\}\; \backslash frac\{(Q\text{'}(t))^2\}\{C\}\; =\; \backslash frac\{1\}\{2\}\; \backslash left(\; \backslash frac\{1\}\{2\}\; \backslash frac\{Q^2\}\{C\}\; \backslash right)$ $\backslash frac\{(Q\text{'}(t))^2\}\{C\}\; =\; \backslash frac\{Q^2\}\{2C\}$ $(Q\text{'}(t))^2\; =\; \backslash frac\{Q^2\}\{2\}$ $Q\text{'}(t)\; =\; \backslash frac\{Q\}\{\backslash sqrt\{2\}\}$ This occurs when $\backslash cos(\backslash omega\_r\; t)\; =\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}$, which corresponds to times $t\; =\; \backslash frac\{T\}\{8\},\; \backslash frac\{3T\}\{8\},\; \backslash frac\{5T\}\{8\},\; \backslash frac\{7T\}\{8\},\; \backslash dots$.
2. If a resistor is inserted in the circuit, the LC circuit becomes an RLC circuit. The oscillations will be damped due to the presence of resistance. The energy stored in the circuit will be gradually dissipated as heat in the resistor due to the Joule heating effect ($I^2R$ losses). In the absence of any external driving force, all the initial energy stored in the capacitor and inductor will eventually be dissipated as heat in the resistor as the oscillations die out. The total amount of energy eventually dissipated as heat will be equal to the total initial energy stored in the circuit, which is 1 Joule.