Top 50 Most Repeated ALTERNATING CURRENT PYQs | JEE MAINS
A curated collection of the most important questions from ALTERNATING CURRENT, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from ALTERNATING CURRENT, fully solved with step-by-step concepts to prepare for JEE MAINS.
Wattless current = I sinφ, where sinφ = √(1 - cos²φ) and I = S/V. Use P = VI cosφ to find I....
Read Full Step-by-Step Solution →Input power = 200 10 = 2000 W. Output = 80% of 2000 = 1600 W....
Read Full Step-by-Step Solution →RMS = $\frac{V_0}{\sqrt{2}} = \frac{100}{1.414} \approx 71$ V....
Read Full Step-by-Step Solution →Use $\omega = 1/\sqrt{LC}$ with proper unit conversion....
Read Full Step-by-Step Solution →At resonance, $\omega L = \frac{1}{\omega C}$ ⇒ net reactance zero....
Read Full Step-by-Step Solution →At resonance $\omega_0 = 1/\sqrt{LC}$, the reactive terms cancel and the impedance equals $R$. Converting the peak voltage to RMS ($V_{rms}=V_0/\sqrt{...
Read Full Step-by-Step Solution →For a sinusoidal waveform, $V_{\text{peak}} = \sqrt{2}\,V_{\text{rms}}$. Substituting $V_{\text{rms}} = 120\ \text{V}$ gives $V_{\text{peak}} = 1.414\...
Read Full Step-by-Step Solution →$ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \approx 141 \Omega $...
Read Full Step-by-Step Solution →Convert $L = 2.0\,\text{mH}=2.0\times10^{-3}\,\text{H}$ and $C = 0.5\,\mu\text{F}=5.0\times10^{-7}\,\text{F}$. Their product $LC = 1.0\times10^{-9}$. ...
Read Full Step-by-Step Solution →Use $ \omega = 1/\sqrt{LC} $. Convert units: $ L=4\times10^{-3}, C=16\times10^{-6} $, then compute $ \omega = 1250\,\text{rad/s} $....
Read Full Step-by-Step Solution →Wattless current = $I_{rms} \sin\phi$, $\cos\phi = 0.6 → \sin\phi = 0.8$...
Read Full Step-by-Step Solution →Output power = 1800 W. Secondary voltage = 400 V. $I = P/V = 4.5$ A → corrected: $I = 9$ A due to turns ratio and power....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Alternating Current....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Alternating Current....
Read Full Step-by-Step Solution →Wattless current = I sinφ, where sinφ = √(1 - cos²φ) and I = S/V. Use P = VI cosφ to find I....
Read Full Step-by-Step Solution →Input power = 200 10 = 2000 W. Output = 80% of 2000 = 1600 W....
Read Full Step-by-Step Solution →RMS = $\frac{V_0}{\sqrt{2}} = \frac{100}{1.414} \approx 71$ V....
Read Full Step-by-Step Solution →Use $\omega = 1/\sqrt{LC}$ with proper unit conversion....
Read Full Step-by-Step Solution →At resonance, $\omega L = \frac{1}{\omega C}$ ⇒ net reactance zero....
Read Full Step-by-Step Solution →At resonance $\omega_0 = 1/\sqrt{LC}$, the reactive terms cancel and the impedance equals $R$. Converting the peak voltage to RMS ($V_{rms}=V_0/\sqrt{...
Read Full Step-by-Step Solution →For a sinusoidal waveform, $V_{\text{peak}} = \sqrt{2}\,V_{\text{rms}}$. Substituting $V_{\text{rms}} = 120\ \text{V}$ gives $V_{\text{peak}} = 1.414\...
Read Full Step-by-Step Solution →$ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \approx 141 \Omega $...
Read Full Step-by-Step Solution →Convert $L = 2.0\,\text{mH}=2.0\times10^{-3}\,\text{H}$ and $C = 0.5\,\mu\text{F}=5.0\times10^{-7}\,\text{F}$. Their product $LC = 1.0\times10^{-9}$. ...
Read Full Step-by-Step Solution →Use $ \omega = 1/\sqrt{LC} $. Convert units: $ L=4\times10^{-3}, C=16\times10^{-6} $, then compute $ \omega = 1250\,\text{rad/s} $....
Read Full Step-by-Step Solution →Wattless current = $I_{rms} \sin\phi$, $\cos\phi = 0.6 → \sin\phi = 0.8$...
Read Full Step-by-Step Solution →Output power = 1800 W. Secondary voltage = 400 V. $I = P/V = 4.5$ A → corrected: $I = 9$ A due to turns ratio and power....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Alternating Current....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Alternating Current....
Read Full Step-by-Step Solution →Wattless current = I sinφ, where sinφ = √(1 - cos²φ) and I = S/V. Use P = VI cosφ to find I....
Read Full Step-by-Step Solution →Input power = 200 10 = 2000 W. Output = 80% of 2000 = 1600 W....
Read Full Step-by-Step Solution →RMS = $\frac{V_0}{\sqrt{2}} = \frac{100}{1.414} \approx 71$ V....
Read Full Step-by-Step Solution →Use $\omega = 1/\sqrt{LC}$ with proper unit conversion....
Read Full Step-by-Step Solution →At resonance, $\omega L = \frac{1}{\omega C}$ ⇒ net reactance zero....
Read Full Step-by-Step Solution →At resonance $\omega_0 = 1/\sqrt{LC}$, the reactive terms cancel and the impedance equals $R$. Converting the peak voltage to RMS ($V_{rms}=V_0/\sqrt{...
Read Full Step-by-Step Solution →For a sinusoidal waveform, $V_{\text{peak}} = \sqrt{2}\,V_{\text{rms}}$. Substituting $V_{\text{rms}} = 120\ \text{V}$ gives $V_{\text{peak}} = 1.414\...
Read Full Step-by-Step Solution →$ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \approx 141 \Omega $...
Read Full Step-by-Step Solution →Convert $L = 2.0\,\text{mH}=2.0\times10^{-3}\,\text{H}$ and $C = 0.5\,\mu\text{F}=5.0\times10^{-7}\,\text{F}$. Their product $LC = 1.0\times10^{-9}$. ...
Read Full Step-by-Step Solution →Use $ \omega = 1/\sqrt{LC} $. Convert units: $ L=4\times10^{-3}, C=16\times10^{-6} $, then compute $ \omega = 1250\,\text{rad/s} $....
Read Full Step-by-Step Solution →