Using a variable-frequency a.c. voltage source, the maximum current measured in the given LCR circuit is 50 mA for V = 5sin(100t)$V = 5\sin(100t)$. The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ____ mutextF$\mu\text{F}$.

### 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 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

### Related Formula For a series RC branch, the voltage phase angle theta$\theta$ is: tantheta = fracX_CR = frac1omega C R$$\tan\theta = \frac{X_C}{R} = \frac{1}{\omega C R}$$ If the phase difference between V_textin$V_{\text{in}}$ and (V_B - V_A)$(V_B - V_A)$ is 90^circ$90^\circ$: theta = 45^circ implies tantheta = 1$$\theta = 45^\circ \implies \tan\theta = 1$$ ### Core Logic

From the phasor representation of the symmetric RC divider bridge: - The phase angle of V_A$V_A$ is theta$\theta$ behind the input voltage. - The phase angle of V_B$V_B$ is 90^circ - theta$90^\circ - \theta$ ahead of the input. - If their combined relative phase difference is 90^circ$90^\circ$, this symmetry dictates:

theta + theta = 90^circ implies theta = 45^circ$$\theta + \theta = 90^\circ \implies \theta = 45^\circ$$ Therefore, we have the condition: X_C = R implies frac1omega C = R$$X_C = R \implies \frac{1}{\omega C} = R$$ ### Step 1: Solve for Frequency Rearrange to solve for omega$\omega$: omega = frac1R C$$\omega = \frac{1}{R C}$$ Substitute the given values: - R = 100 mathrm~kOmega = 10^5 Omega$R = 100 \mathrm{~k}\Omega = 10^5 \Omega$ - C = 100 mathrm~pF = 100 times 10^-12 mathrm~F = 10^-10 mathrm~F$C = 100 \mathrm{~pF} = 100 \times 10^{-12} \mathrm{~F} = 10^{-10} \mathrm{~F}$ omega = frac110^5 times 10^-10 = frac110^-5 = 10^5 mathrm~rad/sec$$\omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5 \mathrm{~rad/sec}$$ Since omega = 10^x$\omega = 10^x$, we get x = 5$x = 5$. ### Pattern Recognition Sees: Phase shift of 90^circ$90^\circ$ across a symmetric RC bridge. Shortcut: Clamping phase shift at 90^circ$90^\circ$ implies the reactive impedance equals the resistive impedance (X_C = R$X_C = R$). The frequency is simply the characteristic time constant frequency omega = 1/tau = 1/(RC)$\omega = 1/\tau = 1/(RC)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula For a current having both DC and AC components i = I_textdc + I_0 cos(omega t + phi)$i = I_{\text{dc}} + I_0 \cos(\omega t + \phi)$, the mean-square value is: langle i^2 rangle = I_textdc^2 + fracI_0^22$$\langle i^2 \rangle = I_{\text{dc}}^2 + \frac{I_0^2}{2}$$ The RMS current I_textrms$I_{\text{rms}}$ is: I_textrms = sqrtlangle i^2 rangle = sqrtI_textdc^2 + fracI_0^22$$I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \sqrt{I_{\text{dc}}^2 + \frac{I_0^2}{2}}$$ ### Core Logic Identify the parameters from the given equation: - I_textdc = 5sqrt2 mathrm~A$I_{\text{dc}} = 5\sqrt{2} \mathrm{~A}$ - I_0 = 10 mathrm~A$I_0 = 10 \mathrm{~A}$ Calculate the square of the components: I_textdc^2 = (5sqrt2)^2 = 50$$I_{\text{dc}}^2 = (5\sqrt{2})^2 = 50$$ fracI_0^22 = frac1002 = 50$$\frac{I_0^2}{2} = \frac{100}{2} = 50$$ ### Step 1: Compute Resultant RMS Substitute back into the RMS formula: I_textrms = sqrt50 + 50 = sqrt100 = 10 mathrm~Amp$$I_{\text{rms}} = \sqrt{50 + 50} = \sqrt{100} = 10 \mathrm{~Amp}$$ ### Pattern Recognition Sees: Superposition of a DC current I_textdc$I_{\text{dc}}$ and a pure AC cosine wave with amplitude I_textac$I_{\text{ac}}$. Shortcut: Use the orthogonal component RMS formula: I_textrms = sqrtI_textdc^2 + I_textac,rms^2$I_{\text{rms}} = \sqrt{I_{\text{dc}}^2 + I_{\text{ac,rms}}^2}$. Here I_textdc = 5sqrt2$I_{\text{dc}} = 5\sqrt{2}$ and I_textac,rms = 10/sqrt2 = 5sqrt2$I_{\text{ac,rms}} = 10/\sqrt{2} = 5\sqrt{2}$. Thus, I_textrms = sqrt50 + 50 = 10 mathrm~A$I_{\text{rms}} = \sqrt{50 + 50} = 10 \mathrm{~A}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current