Alternating Current Class 12 Important Questions Physics

Ans. capacitative reactance $X_{C}=1 / \omega C=1 / 2 \pi v C$. At very high frequencies $X_{C}=0$

Ans. $\left[M^{0} L^{0} T^{-1}\right]$

Ans. rms value of the ac source is $E_{m s}=E_{0} / \sqrt{2}$

Ans. Since $V_{L}=V_{C}$ , so circuit is resonant and in resonant condition, RLC series circuit behaves like a resistive circuit. Hence phase difference between voltage and current is $\phi=0^{\circ}$

Ans. 10 W

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Ans. Yes, the voltage drops across the inductor or the capacitor in a series circuit can be grater thenthe applied voltage. This is because these voltages are not in phase and they cannot be added like ordinary numbers.

Ans. The resistance is the opposition offered by a resistor R to the flow of ac in a circuit, the reactance is the opposition offered by an inductor L or by a capacitor C or by both while impedance is the opposition offered by L-R, C-R or L-C-R circuits. The resistance is independent of the frequency while the reactance and the impedance do depend on the frequency of ac

Ans. It is clear from the figure

Ans. (i) On reducing the number of turns in the inductor, the inductance $L\left(\propto N^{2}\right)$ and hence the inductive reactance $X_{L}(=\omega L)$ decreases. So, the current in the circuit increases and the bulbs glows more brightly. (ii) When $X_{L}=X_{C}$ the impedance of the circuit is minimum (equal to R). Hence the current is maximum and the bulb glows with maximum brightness

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Ans. At later times, energy is shared between C and L but total energy remains the same, assuming that there is no loss of energy.

Ans. When the frequency of the supply equals the natural frequency of the circuit, resonance occurs.

Ans. The emf applied across the inductance L is $E=E_{0}$ $\sin \omega t$ If $\frac{d i}{d t}$ rate of change of current through L at any instant, then induced emf in the inductor at the same instant is $e=-L \frac{d i}{d t}$ To maintain the flow of current in this circuit, applied voltage must be equal and opposite to the

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Ans. When a source of ac is connected to an inductor L, current in it grows from zero to maximum steady value $I_{0}$ An induced emf developes in the inductor, which opposes the growth of current

Ans. Suppose an alternating emf $E=E_{0} \sin \omega t$ is applied across an ideal capacitor then current flows through the circuit is $i=i_{0} \sin (\omega t+\pi / 2$ $(\because$In purely capacitive circuits, current leads the voltage by 90°) Hence, work done over a complete cycle of ac is Hence, work done over a complete cycle of ac is

Ans. The opposition offered by series RLC circuit to the flow of current through it is called impedance of the circuit. It is denoted by Z. Suppose a resistance R, inductor L and capacitor of capacitance C are connected in series to an alternating voltage supply $V=V_{0} \sin \omega t$ as shown The voltage across resistance R is given by $V_{R}=i R$ it is in phase with current and it is denoted by vector $\overrightarrow{\mathrm{OA}}$ (ii) The voltage across inductor L is given by $V_{L}=i X_{L}$ it leads the current by $90^{\circ}$and it is represented by vector $\overrightarrow{O B}$ along Y (iii) The voltage across capacitor C is given by (iii) The voltage across capacitor C is given by it lag behind the current by 90°, and it is represented by along it lag behind the current by $90^{\circ}$, and it is represented by $\overrightarrow{O C}$ along $Y^{\prime}$ As the voltage across $L$ and $C$ have phase difference by $180^{\circ}$ the voltage across $L C .$ Combination is $\left(V_{L}-V_{c}\right),$ assuming that $V_{L}>V_{C}$ It is represented by vector $O B$ in phasor diagram the resultant of $\frac{1}{O A}$ and $\overline{O B}$ is along $\overline{O K}$ hence from phasor diagram it is clear that $|$

Ans. A transformer have following losses. Copper loss : This loss is due to heating of primary and secondary copper coils of transformer. This loss can be minimised by using thick Cu wires in windings. Iron loss : When a changing magnetic flux is linked with the iron core of the transformer, eddy currents are set up in it. These eddy currents produces heat energy (i.e. loss of energy). The energy loss is minimised by using laminated iron cores. Hysteresis loss : This loss occurs due to the repeated magnetisitaion and demagnetisation of iron core when ac is fed to it. This loss can be minimised by using a core of material like soft iron which has a low hysthersis loss. Magnetic flux leakage : Rate of change of magentic flux linked with the primary coil may not be exactly equal to the rate of change of magnetic flux linked with the secondary coil. It can be minimised by winding the secondary coil inside the primary coil. Vibration of core : A transformer produces humming noise due to magnetisation effect. Some energy is lost in the form of mechanical energy to produce vibration in the core.

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Ans. (i) Source frequency in resonance In resonance $V_{C}=V_{L}=1437.5 \mathrm{V}$ The algebraic sum of the three voltages is more than the source voltage of 230 V. These voltages are not in the same phase and cannot be added like ordinary numbers. The voltage across L and C are out of phase and get added to zero. So $\left(V_{R}\right)_{m s}=$ applied rms voltage.

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