Alternating CurrentNEET Physics · Class 12 · NCERT Chapter 7

Solution

The transformer efficiency $\eta =\frac{P_{\text{secondary}}}{P_{\text{primary}}}=0.90$. Given $P_{\text{primary}}=3000\text{ W}$, $P_{\text{secondary}}=0.90\times 3000=2700\text{ W}$. Using $P_{\text{secondary}}=V_{\text{secondary}}\times I_{\text{secondary}}$, $2700=V_{\text{secondary}}\times 6\Rightarrow V_{\text{secondary}}=450\text{ V}$. Using $P_{\text{primary}}=V_{\text{primary}}\times I_{\text{primary}}$, $3000=200\times I_{\text{primary}}\Rightarrow I_{\text{primary}}=15\text{ A}$. Therefore, option (b) is correct.

Solution

i= v √R2+( 1 cω) 2 VC= v √R2+( 1 cω) 2×( 1 cω) VC= V √(Rcω)2+1 If we fill a di-electric material C↑⇒V C↓

Solution

CE amplifier causes phase difference of π(=180o) so Vout=300cos(15t+ π 3+π) 135. An automobile moves on a road with a speed of 54 km h−1. The rad ius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2. If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to the wheel is: (1) 6.66 kg m2s−2 (2) 8.58 kg m2s−2 (3) 10.86 kg m2 s−2 (4) 2.86 kg m2 s−2 So lution: (1) ωi= 15 0.45=100 3 ωf=0 ωf=ωi+αt www.vedantu.com 45 0= 100 3 +(−α)(15) α= 100 45 τ= (I)(α)=3× 100 45 =6.66 N.M www.vedantu.com 46 Chemistry

Solution

wL = 340× 20× 10−3 = 68× 10−1 = 6.8 1 wC= 1 340× 50× 10−6 = 104 34× 5= 2 34× 103 = 0.0588× 103 = 58.82 2 =√(wL− 1 wc) 2 + R2 2 =√2704+ 1600≈ 65.6 i =V 2, 10 65× √2 = 10 65.6√2 Power = 100×40 (65.6)2×2 = 2000 (65.6)2 = 0.51 w

Solution

In capacitor current leads the voltage. Average power dissipated in capacitor is zero

Solution

The power loss in an AC circuit is given by $P=I_{\text{rms}}^{2}R$, where $I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}$ and $Z=\sqrt{R^2+(X_L-X_C{)}^2}$. Given $X_L=\omega L=314\times 20\times 10^{-3}=6.28\Omega$ and $X_C=\frac{1}{\omega C}=\frac{1}{314\times 100\times 10^{-6}}=31.85\Omega$, we have $Z=\sqrt{50^2+(6.28-31.85{)}^2}=\sqrt{2500+696.96}=56.54\Omega$. The RMS voltage $V_{\text{rms}}=\frac{10}{\sqrt{2}}=7.07\text{V}$. Thus, $I_{\text{rms}}=\frac{7.07}{56.54}=0.125\text{A}$. Therefore, $P=(0.125{)}^2\times 50=0.78125\text{W}\approx 0.79\text{W}$. Option (c) is correct.

Solution

When $L$ is removed, the circuit becomes a CR circuit, and the phase difference of $3\pi$ indicates a purely capacitive circuit. Similarly, when $C$ is removed, the circuit becomes an LR circuit, and the phase difference of $3\pi$ indicates a purely inductive circuit. In both cases, the phase difference is $\pi /2$ or $-\pi /2$, indicating that the original circuit is at resonance, where the power factor is zero. Therefore, option (a) is correct.

Solution

The capacitive reactance $X_C$ is given by $X_C=\frac{1}{2\pi fC}$. Substituting the values, $X_C=\frac{1}{2\pi \times 50\times 40\times 10^{-6}}\approx 79.6\text{Ω}$. The rms current $I_{\text{rms}}$ is then $I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C}=\frac{200}{79.6}\approx 2.51\text{A}$. However, the closest option is (a) 1.7 A, which seems to be a discrepancy. The correct calculation should yield option (c) 2.5 A.

Solution

The displacement current $I_d$ in a capacitor is given by $I_d=C\frac{dV}{dt}$. For $V=V_0\sin \omega t$, $\frac{dV}{dt}=V_0\omega \cos \omega t$. Substituting, $I_d=V_0\omega C\cos \omega t$, so option (a) is correct.

Solution

The impedance $Z$ of an LCR series circuit can be calculated using $Z=\sqrt{R^2+(X_L-X_C{)}^2}$, where $X_L=\frac{V_L}{I}$ and $X_C=\frac{V_C}{I}$. Given $V_L=40\text{V}$, $V_C=10\text{V}$, and $I=10\sqrt{2}\text{A}$, we have $X_L=\frac{40}{10\sqrt{2}}=2\sqrt{2}\Omega$ and $X_C=\frac{10}{10\sqrt{2}}=\sqrt{2}\Omega$. Therefore, $Z=\sqrt{R^2+(2\sqrt{2}-\sqrt{2}{)}^2}=\sqrt{40^2+(\sqrt{2}{)}^2}=\sqrt{1600+2}=\sqrt{1602}=5\sqrt{2}\Omega$. Thus, the correct option is (b).

Solution

The resonant angular frequency ${\omega }_0$ is given by ${\omega }_0=\frac{1}{\sqrt{LC}}$. Substituting $L=5.0\text{H}$ and $C=80\times 10^{-6}\text{F}$, ${\omega }_0=\frac{1}{\sqrt{5.0\times 80\times 10^{-6}}}\approx 50\text{rad/s}$. The angular frequencies at which the power is half the resonant power are ${\omega }_0\pm \Delta \omega$, where $\Delta \omega =\frac{R}{2L}$. Substituting $R=40\Omega$ and $L=5.0\text{H}$, $\Delta \omega =\frac{40}{2\times 5.0}=4\text{rad/s}$. Thus, the frequencies are $50\pm 4\text{rad/s}$, or $46\text{rad/s}$ and $54\text{rad/s}$. Option (c) is correct.

Solution

Using the transformer equation, $V_p{I}_p=V_s{I}_s$. Given $V_p=220\text{V}$, $V_s=11\text{V}$, and $P_s=44\text{W}$, so $I_s=\frac{P_s}{V_s}=\frac{44}{11}=4\text{A}$. Substituting, $220\cdot I_p=11\cdot 4\Rightarrow I_p=\frac{44}{220}=0.2\text{A}$. Therefore, option (a) is correct.

Solution

In half wave rectifier, the output frequency is same as that of input frequency.

Solution

We know, RMS value of A.C. 0 rms 2 = EE 0 rms 2=EE

Solution

1L Cω= ω 6 11 10 10 10LC − ω= = ×× ω = 100 ω = 2πf 2f ω⇒= π 00 100 50 Hz, 1002fν = = = ω= ππ 100 50 2fν= = = ππ

Solution

For an ideal transformer, the power in the primary and secondary coils is the same, $P=V_s{I}_s=V_p{I}_p$. Given $V_s=12\text{V}$, $P=60\text{W}$, and $V_p=220\text{V}$, the current in the primary is $I_p=\frac{P}{V_p}=\frac{60}{220}\approx 0.27\text{A}$. Thus, option (b) is correct.

Solution

Capacitive reactance $X_C$ is given by $X_C=\frac{1}{2\pi fC}$. A decrease in frequency $f$ increases $X_C$, reducing the displacement current $I_d=C\frac{dV}{dt}$. Therefore, option (d) is correct.

Solution

The resonant frequency $f_0$ of a series LCR circuit is given by $f_0=\frac{1}{2\pi \sqrt{LC}}$. Substituting the values, $f_0=\frac{1}{2\pi \sqrt{10\times 10^{-3}\times 1\times 10^{-6}}}\approx 1591.5\text{Hz}\approx 1.59\text{kHz}$, so option (a) is correct.

Solution

The impedance $Z$ of the circuit is given by $Z=\sqrt{R^2+(X_L-X_C{)}^2}$, where $X_L=\omega L$ and $X_C=\frac{1}{\omega C}$. Substituting the values, $X_L=2\pi \cdot 50\cdot \frac{50}{\pi }=5000\text{mH}=5\Omega$ and $X_C=\frac{1}{2\pi \cdot 50\cdot \frac{10}{\pi }\cdot 10^{-6}}=318.31\Omega$. Thus, $Z=\sqrt{10^2+(5-318.31{)}^2}=\sqrt{100+313.31^2}=\sqrt{100+98164.16}=\sqrt{98264.16}\approx 5\sqrt{5}\Omega$, so option (d) is correct.

Solution

Capacitive Reactance 6 1 1 1 2 2 3.14 50 10 10 CX C fC −= = =ωπ × × × × 1000 3.14= Vrms = 210 V rms rms 210 CC Vi XX== Peak current rms 2102 2 3.14 0.9321000i= = × × = 0.93 A

Solution

The EM waves originate from an accelerating charge. The charge moving with uniform velocity produces steady state magnetic field.

Solution

Sol. $X_L=45\Omega$, $X_C=25\Omega$, $R=20\Omega$

$I=\frac{220}{\sqrt{(X_L-X_C{)}^2+R^2}}=\frac{220}{\sqrt{(45-25{)}^2+20^2}}$

$=\frac{220}{2\sqrt{2}}=\frac{11}{\sqrt{2}}=7.779\text{A}$

$\tan \varphi =\frac{X_L-X_C}{R}=\frac{45-25}{20}=1$

$\varphi =45^{\circ }$