An inductor stores 16 text J$16 \text{ J}$ of magnetic field energy and dissipates 32 text W$32 \text{ W}$ of thermal energy due to its resistance when an a.c. current of 2 text A$2 \text{ A}$ (rms) and frequency 50 text Hz$50 \text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is ____. ( pi = 3.14$\pi = 3.14$ )

### Related Formula omega = frac1sqrtLC$$\omega = \frac{1}{\sqrt{LC}}$$ ### Core Logic Current in an LCR circuit reaches a maximum value (measured as 50 text mA$50 \text{ mA}$) exactly when the circuit is in resonance. At resonance, the inductive reactance equals the capacitive reactance, meaning X_L = X_C$X_L = X_C$, which gives the condition omega = 1/sqrtLC$\omega = 1/\sqrt{LC}$. ### Step 1: Identify Parameters From the source equation V = 5sin(100t)$V = 5\sin(100t)$: omega = 100 text rad/s$\omega = 100 \text{ rad/s}$ From the circuit diagram: L = 2 text H$L = 2 \text{ H}$ ### Step 2: Evaluate Capacitance omega^2 = frac1LC$$\omega^{2} = \frac{1}{LC}$$ C = frac1omega^2L$$C = \frac{1}{\omega^{2}L}$$ C = frac1(100)^2 times 2 = frac110000 times 2 = frac12 times 10^4$$C = \frac{1}{(100)^{2} \times 2} = \frac{1}{10000 \times 2} = \frac{1}{2 \times 10^{4}}$$ C = 50 times 10^-6 text F$$C = 50 \times 10^{-6} \text{ F}$$ C = 50 mutextF$$C = 50 \mu\text{F}$$ ### Pattern Recognition Sees: "maximum current" + "variable-frequency / given frequency" → Resonance! The impedance is purely resistive (Z=R$Z=R$), so omega L = 1/(omega C)$\omega L = 1/(\omega C)$. The 50 text mA$50 \text{ mA}$ info is a distractor/cross-check (since V_peak/R = 5/100 = 50 text mA$V_{peak}/R = 5/100 = 50 \text{ mA}$, confirming resonance). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula textImpedance, Z = sqrtR^2 + (X_L - X_C)^2$$\text{Impedance, } Z = \sqrt{R^2 + (X_L - X_C)^2}$$ textPower factor, cos phi = fracRZ$$\text{Power factor, } \cos \phi = \frac{R}{Z}$$ ### Core Logic Given from the circuit: Resistance, R = 60\, Omega$R = 60\, \Omega$ Inductive reactance, X_L = 70\, Omega$X_L = 70\, \Omega$ Capacitive reactance, X_C = 150\, Omega$X_C = 150\, \Omega$ First, calculate the impedance Z$Z$: Z = sqrt60^2 + (150 - 70)^2$$Z = \sqrt{60^2 + (150 - 70)^2}$$ Z = sqrt60^2 + 80^2 = sqrt3600 + 6400 = sqrt10000 = 100\, Omega$$Z = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\, \Omega$$ ### Step 1: Calculate Power Factor Power factor = fracRZ$= \frac{R}{Z}$ textPower factor = frac60100 = frac610$$\text{Power factor} = \frac{60}{100} = \frac{6}{10}$$ Given that the power factor is fracalpha10$\frac{\alpha}{10}$, comparing both sides gives alpha = 6$\alpha = 6$. ### Pattern Recognition A classic 3-4-5 impedance triangle where R=60, X_C-X_L=80$R=60, X_C-X_L=80$, immediately implies Z=100$Z=100$. The power factor is exactly R/Z = 60/100 = 6/10$R/Z = 60/100 = 6/10$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula I_mathrmrms = sqrt fracint_0^T i^2 \, dtint_0^T dt $$I_{\mathrm{rms}} = \sqrt{ \frac{\int_{0}^{T} i^2 \, dt}{\int_{0}^{T} dt} }$$ ### Core Logic To find the root mean square (rms) value of a varying current, we integrate the square of the current over the given time period, divide by the time period, and take the square root. ### Step 1: Integration of Squared Current mathrmi_mathrmrms^2 = fracint_0^mathrmT left( mathrmi_0^2 mathrmt^2 / mathrmT^2 right) mathrmdtint_0^mathrmT mathrmdt$$\mathrm{i}_{\mathrm{rms}}^2 = \frac{\int_{0}^{\mathrm{T}} \left( \mathrm{i}_{0}^2 \mathrm{t}^2 / \mathrm{T}^2 \right) \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}$$ = fracmathrmi_0^2mathrmT^3 int_0^mathrmT mathrmt^2 \, mathrmdt$$= \frac{\mathrm{i}_{0}^2}{\mathrm{T}^3} \int_{0}^{\mathrm{T}} \mathrm{t}^2 \, \mathrm{dt}$$ ### Step 2: Final Calculation mathrmi_mathrmrms^2 = fracmathrmi_0^2mathrmT^3 cdot left[ fracmathrmt^33 right]_0^mathrmT$$\mathrm{i}_{\mathrm{rms}}^2 = \frac{\mathrm{i}_{0}^2}{\mathrm{T}^3} \cdot \left[ \frac{\mathrm{t}^3}{3} \right]_{0}^{\mathrm{T}}$$ = fracmathrmi_0^2mathrmT^3 cdot fracmathrmT^33 = fracmathrmi_0^23$$= \frac{\mathrm{i}_{0}^2}{\mathrm{T}^3} \cdot \frac{\mathrm{T}^3}{3} = \frac{\mathrm{i}_{0}^2}{3}$$ i_mathrmrms = fraci_0sqrt3$$i_{\mathrm{rms}} = \frac{i_{0}}{\sqrt{3}}$$ ### Pattern Recognition For any linearly varying quantity y = kt$y = kt$ passing through origin, the RMS value over time T$T$ is always fractextMaximum Valuesqrt3$\frac{\text{Maximum Value}}{\sqrt{3}}$. A standard AC sine wave is 1/sqrt2$1/\sqrt{2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula Power consumed in a purely resistive AC device (like a lightbulb) is: P = V_textrms I_textrms$$P = V_{\text{rms}} I_{\text{rms}}$$ The peak current I_0$I_0$ is related to the rms current I_textrms$I_{\text{rms}}$ by: I_0 = sqrt2 I_textrms$$I_0 = \sqrt{2} I_{\text{rms}}$$ ### Core Logic Given parameters: - Power P = 100mathrm~W$P = 100\mathrm{~W}$ - rms Voltage V_textrms = 220mathrm~V$V_{\text{rms}} = 220\mathrm{~V}$ ### Step 1: Calculate RMS Current (I_textrms$I_{\text{rms}}$) I_textrms = fracPV_textrms = frac100220 = frac511mathrm~A approx 0.455mathrm~A$$I_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{100}{220} = \frac{5}{11}\mathrm{~A} \approx 0.455\mathrm{~A}$$ ### Step 2: Calculate Peak Current ($ ### Step 2: Calculate Peak Current ($I_0) $) $I_0 = sqrt2 I_textrms = 1.414 times frac511 = frac7.0711 approx 0.64mathrm~A$I_0 = \sqrt{2} I_{\text{rms}} = 1.414 \times \frac{5}{11} = \frac{7.07}{11} \approx 0.64\mathrm{~A}$ ### Pattern Recognition Always remember that rated values specify the RMS limits. Since the source voltage matches the bulb's rated voltage exactly, the actual power equals the rated power ($ ### Pattern Recognition Always remember that rated values specify the RMS limits. Since the source voltage matches the bulb's rated voltage exactly, the actual power equals the rated power ($100\mathrm{~W}). Do not confuse $). Do not confuse $I_{\text{rms}} ($ ($0.45\mathrm{~A}) with the peak current $) with the peak current $I_0 ($ ($0.64\mathrm{~A}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current