Alternating CurrentNEET Physics · Class 12 · NCERT Chapter 7

Solution

$i_{\text{rms}}=i_0/\sqrt{2}$.

Solution

In an inductor, V leads I by π/2.

Solution

In a capacitor, I leads V by π/2.

Solution

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$.

Solution

$f_0=1/(2\pi \sqrt{LC})$.

Solution

$V_p/V_s=N_p/N_s$.

Solution

Pure L: φ = π/2, cos φ = 0. Average power = 0.

Solution

$V_0=V_{\text{rms}}\sqrt{2}=220\times 1.414\approx 311\text{V}$.

Solution

$X_L=2\pi fL=2\pi \times 50\times 0.05\approx 15.7\Omega$.

Solution

$X_C=1/(2\pi fC)=1/(2\pi \times 50\times 100\times 10^{-6})\approx 31.8\Omega$.

Solution

At resonance, X_L = X_C, so V and I are in phase: φ = 0.

Solution

$N_s/N_p=V_s/V_p=11000/220=50$.

Solution

Pure R: φ = 0, cos φ = 1.

Solution

Average of sin over a full cycle is zero.

Solution

$Q=\frac{1}{R}\sqrt{L/C}$.

Solution

Z is min (= R) at resonance, so I is max (= V/R).

Solution

P loss = I² R. For fixed P, higher V means lower I, drastically reducing line losses.

Solution

Pure L: cos φ = 0. Average power = 0.

Solution

X_C = 1/(ωC) → 0 at high f. Capacitor passes high frequencies.

Solution

$V_0=220\sqrt{2}\approx 311\text{V}$, $\omega =2\pi \times 50=100\pi$ rad/s.

Solution

$Z=\sqrt{64+36}=10\Omega$.

Solution

Transformer needs CHANGING flux. DC produces constant flux: no induction.

Solution

$f_0=1/(2\pi \sqrt{LC})=1/(2\pi \times 10^{-3})\approx 159\text{Hz}$.

Solution

$P=V_{\text{rms}}{I}_{\text{rms}}\cos 60°=100\times 4\times 0.5=200\text{W}$.

Solution

$X_L=2\pi fL\Rightarrow f=X_L/(2\pi L)=10/(2\pi \times 0.1)\approx 15.9\text{Hz}$.

Solution

$i_{\text{avg, half}}=2i_0/\pi$.

Solution

$\tan \varphi =(X_L-X_C)/R$. Positive when $X_L>X_C$.

Solution

$I_s/I_p=N_p/N_s=1/10$. $I_s=0.5\text{A}$.

Solution

$X_C=1/(2\pi \times 50\times 10\times 10^{-6})=318\Omega$. $I_{\text{rms}}=220/318\approx 0.69\text{A}$.

Solution

At resonance, X_L = X_C and currents are equal. So V_L = I X_L = I X_C = V_C.

Solution

Wattless current is the part of I 90° out of phase with V; it does no real work over a cycle.

Solution

Ideal transformer: $P_p=P_s$. Voltage drops, current rises in step-down.