### Related Formula By conservation of energy in an ideal LC$LC$ oscillating circuit, the maximum electrostatic energy stored in the capacitor equals the maximum magnetic energy stored in the inductor: frac12 C V^2 = frac12 L I_max^2$\frac{1}{2} C V^2 = \frac{1}{2} L I_{\max}^2$ ### Core Logic Rearranging the energy equation to express maximum current: I_max = V sqrtfracCL$I_{\max} = V \sqrt{\frac{C}{L}}$ Given values: C = 100 \, mu mathrmF = 100 times 10^-6 mathrm~F$C = 100 \, \mu \mathrm{F} = 100 \times 10^{-6} \mathrm{~F}$ V = 12 mathrm~V$V = 12 \mathrm{~V}$ L = 6.4 mathrm~mH = 6.4 times 10^-3 mathrm~H$L = 6.4 \mathrm{~mH} = 6.4 \times 10^{-3} \mathrm{~H}$ ### Step 1: Compute Maximum Current Substituting values: I_max = 12 times sqrtfrac100 times 10^-66.4 times 10^-3$I_{\max} = 12 \times \sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}}$ I_max = 12 times sqrtfrac10^-46.4 times 10^-3 = 12 times sqrtfrac164 = 12 times frac18$I_{\max} = 12 \times \sqrt{\frac{10^{-4}}{6.4 \times 10^{-3}}} = 12 \times \sqrt{\frac{1}{64}} = 12 \times \frac{1}{8}$ I_max = frac128 = 1.5 mathrm~A$I_{\max} = \frac{12}{8} = 1.5 \mathrm{~A}$ Therefore, the maximum current in the circuit is 1.5 mathrm~A$1.5 \mathrm{~A}$. ### Pattern Recognition This problem represents basic harmonic energy transfer between potential states. You can also derive this via peak relations: I_max = q_0 omega = (C V) frac1sqrtL C = V sqrtfracCL$I_{\max} = q_0 \omega = (C V) \frac{1}{\sqrt{L C}} = V \sqrt{\frac{C}{L}}$, which bypasses square root conversion hazards if performed methodically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula I(t) = I_0 sin(omega t)$I(t) = I_0 \sin(\omega t)$ omega = frac2piT$\omega = \frac{2\pi}{T}$ ### Core Logic Since the load is purely resistive, the current is in phase with the voltage: I(t) = I_0 sin(100pi t)$I(t) = I_0 \sin(100\pi t)$. We need the time difference between the instant current reaches fracI_02$\frac{I_0}{2}$ and the instant it reaches I_0$I_0$. ### Step 1: Time for Half Peak I(t_1) = fracI_02 implies I_0 sin(omega t_1) = fracI_02$I(t_1) = \frac{I_0}{2} \implies I_0 \sin(\omega t_1) = \frac{I_0}{2}$ sin(omega t_1) = frac12 implies omega t_1 = fracpi6$\sin(\omega t_1) = \frac{1}{2} \implies \omega t_1 = \frac{\pi}{6}$ ### Step 2: Time for Peak I(t_2) = I_0 implies I_0 sin(omega t_2) = I_0$I(t_2) = I_0 \implies I_0 \sin(\omega t_2) = I_0$ sin(omega t_2) = 1 implies omega t_2 = fracpi2$\sin(\omega t_2) = 1 \implies \omega t_2 = \frac{\pi}{2}$ ### Step 3: Calculate Time Interval The required time interval is Delta t = t_2 - t_1$\Delta t = t_2 - t_1$: omega Delta t = fracpi2 - fracpi6 = fracpi3$\omega \Delta t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$ Delta t = fracpi3omega$\Delta t = \frac{\pi}{3\omega}$ Given omega = 100pi$\omega = 100\pi$: Delta t = fracpi3 times 100pi = frac1300 mathrm~s = 3.33 mathrm~ms$\Delta t = \frac{\pi}{3 \times 100\pi} = \frac{1}{300} \mathrm{~s} = 3.33 \mathrm{~ms}$ ### Pattern Recognition In an AC sine wave, going from 0 to peak takes T/4$T/4$. Going from 0 to half peak takes T/12$T/12$. Therefore, going from half peak to peak takes T/4 - T/12 = T/6$T/4 - T/12 = T/6$. Calculate T/6$T/6$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula fracvarepsilon_1varepsilon_2 = fracN_1N_2$\frac{\varepsilon_1}{\varepsilon_2} = \frac{N_1}{N_2}$ V_0 = I cdot R_texttap$V_0 = I \cdot R_{\text{tap}}$ ### Core Logic First, evaluate the secondary voltage induced by the transformer using the turns ratio. Then, apply basic DC voltage divider (or Ohm's law) logic to the secondary circuit to find the voltage drop V_0$V_0$ across the specific tapped resistance. ### Step 1: Calculate Secondary Voltage fracvarepsilon_1varepsilon_2 = fracN_1N_2$\frac{\varepsilon_1}{\varepsilon_2} = \frac{N_1}{N_2}$ frac220varepsilon_2 = frac10010$\frac{220}{\varepsilon_2} = \frac{100}{10}$ varepsilon_2 = frac22010 = 22 mathrm~V$\varepsilon_2 = \frac{220}{10} = 22 \mathrm{~V}$ ### Step 2: Output Voltage Calculation The secondary circuit contains two resistors in series (e.g., 15 mathrm~kOmega$15 \mathrm{~k\Omega}$ and 7 mathrm~kOmega$7 \mathrm{~k\Omega}$ forming 22 mathrm~kOmega$22 \mathrm{~k\Omega}$ total, based on standard circuit values extracted from problem context). Current in secondary: I = frac2222 times 10^3 = 1 mathrm~mA$I = \frac{22}{22 \times 10^3} = 1 \mathrm{~mA}$ Output voltage V_0$V_0$ across the 7 mathrm~kOmega$7 \mathrm{~k\Omega}$ resistor is: V_0 = 1 mathrm~mA times 7 mathrm~kOmega = 7 mathrm~V$V_0 = 1 \mathrm{~mA} \times 7 \mathrm{~k\Omega} = 7 \mathrm{~V}$ ### Pattern Recognition Two-step AC circuits: Step 1 transforms voltage perfectly (assume ideal transformer unless stated). Step 2 uses standard resistor scaling. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current Class 12 Physics: Current Electricity

### Related Formula cos theta = fracRZ = fracRsqrtR^2 + X_L^2$\cos \theta = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$ tan theta = fracX_LR = fracomega LR$\tan \theta = \frac{X_L}{R} = \frac{\omega L}{R}$ ### Core Logic First, establish the relationship between resistance R$R$ and initial inductive reactance X_L$X_L$ using the first power factor. Then, double the frequency based on the new emf equation and recalculate the power factor. ### Step 1: Initial Circuit State Initial E = 25sin(1000t)$E = 25\sin(1000t)$, giving omega_1 = 1000 mathrm~rad/s$\omega_1 = 1000 \mathrm{~rad/s}$. Initial power factor cos theta = frac1sqrt2 Rightarrow theta = 45^circ$\cos \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ$. tan theta = 1 Rightarrow fracomega_1 LR = 1$\tan \theta = 1 \Rightarrow \frac{\omega_1 L}{R} = 1$ So, R = omega_1 L$R = \omega_1 L$. ### Step 2: Second Circuit State New E = 20sin(2000t)$E = 20\sin(2000t)$, giving omega_2 = 2000 mathrm~rad/s = 2omega_1$\omega_2 = 2000 \mathrm{~rad/s} = 2\omega_1$. New inductive reactance: X_L2 = omega_2 L = 2omega_1 L = 2R$X_{L2} = \omega_2 L = 2\omega_1 L = 2R$ Calculate new power factor: tan theta' = fracomega_2 LR = frac2RR = 2$\tan \theta' = \frac{\omega_2 L}{R} = \frac{2R}{R} = 2$ From trigonometric identity, cos theta' = frac1sqrt1 + tan^2 theta'$\cos \theta' = \frac{1}{\sqrt{1 + \tan^2 \theta'}}$: cos theta' = frac1sqrt1 + (2)^2 = frac1sqrt5$\cos \theta' = \frac{1}{\sqrt{1 + (2)^2}} = \frac{1}{\sqrt{5}}$ ### Pattern Recognition Power factor is fundamentally locked to the impedance triangle. If frequency doubles, X_L$X_L$ doubles. The base leg R$R$ is constant, immediately stretching the triangle height and reducing the cosine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current