\u21d2Xc=10\u00d7C=\u221eSo, for a constant DC source, reactance of a capacitor is infinite.<\/p>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
The voltage and current in a series AC circuit are given by V = V0<\/sub>cos \u03c9t<\/em> and i<\/em> = i<\/em>0<\/sub> sin \u03c9t<\/em>. What is the power dissipated in the circuit?<\/p>\n\n\n\n Voltage, V <\/em>= V<\/em>0<\/sub>cos \u03c9t<\/em> \u2013 \u03c02)<\/p>\n\n\n\n Power dissipated in an AC circuit is given by,<\/p>\n\n\n\n P=IrmsVrmscos\u03d5, \u03c0\/2<\/p>\n\n\n\n \u21d2cos\u03d5=cos\u03c02=0\u2234P=IrmsVrms\u00d70=0<\/p>\n\n\n\n Two alternating currents are given by<\/p>\n\n\n\n i1=i0 sin \u03c9t and i2=i0 sin \u03c9t+\u03c03Will the rms values of the currents be equal or different?<\/p>\n\n\n\n The rms value of current is given by,<\/p>\n\n\n\n irms=i02Since peak value of current i0<\/sub><\/em> is same for both currents, their rms values will be same.<\/p>\n\n\n\n Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?<\/p>\n\n\n\n Let a LCR circuit is connected across an AC supply with the emf E<\/em> = E<\/em>0<\/sub> sin \u03c9t<\/em>. Z=R2+(XL-XC)2Where, V=LdidtThe current in the circuit can be written as<\/p>\n\n\n\n I=I0sin(\u03c9t+\u03d5)Where, \u00cf\u2022<\/em> is the phase difference between the current and the supply voltage V=LI0cos(\u03c9t+\u03d5)Thus peak value of the voltage across the inductor is given by<\/p>\n\n\n\n V=LI0\u21d2V=E0Z\u00d7LTherefore, the peak voltage across the inductor is given by<\/p>\n\n\n\n V=E0Z\u00d7LAt resonance Z = R<\/em>,<\/p>\n\n\n\n V=E0R\u00d7L, LR>1V<\/em> > E<\/em>0<\/sub> LR>1at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.<\/p>\n\n\n\n In a circuit, containing a capacitor and an AC source, the current is zero at the instant the source voltage is maximum. Is it consistent with Ohm\u2019s Law?<\/p>\n\n\n\n Ohm\u2019s Law is valid for resistive circuits only. It is not valid for capacitive or inductive circuits, or a combination of both.<\/p>\n\n\n\n An AC source is connected to a capacitor. Will the rms current increase, decrease or remain constant if a dielectric slab is inserted into the capacitor?<\/p>\n\n\n\n The reactance of a capacitor is given by,<\/p>\n\n\n\n Xc=1\u03c9CAlso,<\/p>\n\n\n\n C=K\u03b50Ad, K>1<\/p>\n\n\n\n \u2234The capacitance C<\/em> of the capacitor will increase on inserting the dielectric slab and, consequently, the reactance Xc<\/sub><\/em> will decrease. irms=\u03b502XCTherefore, rms current will decrease.<\/p>\n\n\n\n When the frequency of the AC source in an LCR circuit equals the resonant frequency, the reactance of the circuit is zero. Does it mean that there is no current through the inductor or the capacitor?<\/p>\n\n\n\n The condition for resonance is:<\/p>\n\n\n\n 1\u03c9C=\u03c9LThe peak current through the circuit is given by,<\/p>\n\n\n\n i0=V0R2+1\u03c9C-\u03c9L2From the condition of resonance, we get:<\/p>\n\n\n\n i0=V0RThe current will flow through the all circuit elements. But since the reactance of the capacitor and inductor are equal, the potential difference across them will be equal and opposite and will cancel each other.<\/p>\n\n\n\n When an AC source is connected to a capacitor, there is a steady-state current in the circuit. Does it mean that the charges jump from one plate to the other to complete the circuit?<\/p>\n\n\n\n No. When an AC source is connected to a capacitor, there is a steady in the circuit to transfer change to the plates of the capacitor. This produces a potential difference between the plates. The capacitance is alternatively charged and discharged as the current reverses after each half cycle.<\/p>\n\n\n\n A current i<\/em>1<\/sub> = i<\/em>0<\/sub> sin \u03c9t<\/em> passes through a resistor of resistance R. How much thermal energy is produced in one time period? A current i<\/em>2<\/sub> = \u2212i<\/em>0<\/sub> sin \u03c9t<\/em> passes through the resistor. How much thermal energy is produced in one time period? If i<\/em>1<\/sub> and i<\/em>2<\/sub> both pass through the resistor simultaneously, how much thermal energy is produced? Is the principle of superposition obeyed in this case?<\/p>\n\n\n\n The thermal energy produced for an AC circuit in one time period is given by,<\/p>\n\n\n\n H=Irms2\u00d7R\u00d72\u03c0\u03c9For current, i<\/em>1<\/sub> = i<\/em>0<\/sub> sin \u03c9t<\/em>,<\/p>\n\n\n\n Irms=i02<\/p>\n\n\n\n \u21d2H=i02R2\u00d72\u03c0\u03c9=\u03c0i02R\u03c9For current, i<\/em>2<\/sub> = \u2212i<\/em>0<\/sub> sin \u03c9t<\/em>,<\/p>\n\n\n\n Irms=i02Hence, the same thermal energy will be produced due to this current.<\/p>\n\n\n\n Since, the direction of i<\/em>1<\/sub> and i<\/em>2<\/sub> are opposite and their magnitude is same, the net current through the resistor will become zero when both are passed together. <\/em>Yes, the principle of superposition is obeyed in this case.<\/p>\n\n\n\n Is energy produced when a transformer steps up the voltage?<\/p>\n\n\n\n When a transformer steps up the voltage, the voltage increases but current decreases. Neglecting any loss of energy, the power remains constant and, hence, energy is not produced. It remains constant.<\/p>\n\n\n\n A transformer is designed to convert an AC voltage of 220 V to an AC voltage of 12 V. If the input terminals are connected to a DC voltage of 220 V, the transformer usually burns. Explain.<\/p>\n\n\n\n A transformer is ideally an inductive coil. For an inductor connected across a DC voltage, V-Ldidt=0\u21d2V=Ldidt\u21d2\u222bdi=VL\u222bdt\u21d2i=VtLFor a DC source, the current across the inductor will increase with time and can reach a very large value, which can burn the transformer.<\/p>\n\n\n\n Can you have an AC series circuit in which there is a phase difference of (a) 180\u00b0 (b) 120\u00b0 between the emf and the current?<\/p>\n\n\n\n Let us consider an AC series LCR circuit of angular frequency \u03c9<\/em>. The impedance of the circuit is given by,<\/p>\n\n\n\n Z=R2+\u03c9L-1\u03c9C2The phase difference between V <\/em>and I<\/em> is given by,<\/p>\n\n\n\n tan\u03d5=\u03c9L-1\u03c9CZFrom the above formula, we can clearly see that<\/p>\n\n\n\n \u03d5\u2208-\u03c02,\u03c02So, we cannot have a phase difference of 180\u00b0 or 120\u00b0.<\/p>\n\n\n\n A resistance is connected to an AC source. If a capacitor is included in the series circuit, will the average power absorbed by the resistance increase or decrease? If an inductor of small inductance is also included in the series circuit, will the average power absorbed increase or decrease further?<\/p>\n\n\n\n When a capacitor is included in a series circuit, the impedance of the circuit,<\/p>\n\n\n\n Z=R2+XC2The power absorbed by a resistor is given by,<\/p>\n\n\n\n P=Irms2RSince impedance increases due to introduction of a capacitor, the rms value of current Irms<\/sub><\/em> will decrease and, hence, the power absorbed by the resistor will decrease.<\/p>\n\n\n\n When a small inductance is introduced in the circuit, the impedance of the circuit,<\/p>\n\n\n\n Z=R2+XC-XL2Since the impedance now decreases a little, the rms value of current will increase and, hence, the power absorbed by the resistor will increase.<\/p>\n\n\n\n Can a hot-wire ammeter be used to measure a direct current of constant value? Do we have to change the graduations?<\/p>\n\n\n\n A hot-wire ammeter measures the rms value of current for an alternating current. So, it can be used to measure the direct current of constant value because that constant value will be equal to the rms value of current. As, the rms value of the current is same as the direct current thus we need not change the graduations.<\/p>\n\n\n\n A capacitor acts as an infinite resistance for<\/p>\n\n\n\n (a) DC (a) DC<\/p>\n\n\n\n For a DC source, \u03c9<\/em> = 0. So, the reactance of the capacitance is given by,<\/p>\n\n\n\n XC=1\u03c9C=10=\u221e<\/p>\n\n\n\n An AC source producing emf (a) i<\/em>1<\/sub> > i<\/em>2<\/sub> (c) i<\/em>1<\/sub> < i<\/em>2<\/sub><\/p>\n\n\n\n The charge on the capacitor during steady state is given by,<\/p>\n\n\n\n Q=C\u03b5=\u03b50Ccos100\u03c0s-1t+cos500\u03c0s-1tThe steady state current is, thus, given by,<\/p>\n\n\n\n i=dQdt=\u03b50C\u00d7100\u03c0sin100\u03c0s-1t+\u03b50C\u00d7500\u03c0sin500\u03c0s-1t\u21d2i=100C\u03c0\u03b50 cos100\u03c0s-1t+\u03d51+500C\u03c0\u03b50 cos500\u03c0s-1+\u03d52\u21d2i1=100C\u03c0\u03b50 & i2=500C\u03c0\u03b50\u2234i2>i1<\/p>\n\n\n\n The peak voltage of a 220 V AC source is<\/p>\n\n\n\n (a) 220 V (c) about 310 V<\/p>\n\n\n\n Given: Vp=2\u00d7Vrms=1.414\u00d7220=311 V<\/p>\n\n\n\n An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It<\/p>\n\n\n\n (a) must be zero is 220\/2 V<\/p>\n\n\n\n (b) may be zero<\/p>\n\n\n\n Let the AC voltage be given by,<\/p>\n\n\n\n V=V0sin\u03c9tHere, \u03c9 = 2<\/p>\n\n\n\n \u03c0f = 314 rad\/s Vavg=\u222b00.01Vdt\u222b00.01dt=-V0cos\u03c9t\u03c900.01 =V0\u03c9\u00d70.011-cos\u03c90.01 =V0314\u00d70.011-cos314\u00d70.01 =V03.141-cos\u03c0 =2V0\u03c0=140.127 VAlso, when<\/p>\n\n\n\n V=V0cos\u03c9t,<\/p>\n\n\n\n Vavg=\u222b00.01Vdt\u222b00.01dt=V0sin\u03c9t\u03c900.01 =V0\u03c9\u00d70.01sin\u03c90.01-0 =V0314\u00d70.01sin314\u00d70.01 =V03.14sin\u03c0 =0 The magnetic field energy in an inductor changes from maximum to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is<\/p>\n\n\n\n (a) 20 Hz (b) 50 Hz<\/p>\n\n\n\n The magnetic field energy in an inductor is given by,<\/p>\n\n\n\n E=12Li2The magnetic energy will be maximum when the current will reach its peak value, i0<\/sub><\/em>, and it will be minimum when the current will become zero.<\/p>\n\n\n\n From the above graph of alternating current, we can see that current reduces from maximum to zero in T<\/em>\/4 time, where T<\/em> is the time period. \u21d2T=20 ms\u2234 Frequency, \u03bd = 1T=120\u00d710-3=50 Hz<\/p>\n\n\n\n Which of the following plots may represent the reactance of a series LC combination?<\/p>\n\n\n\n Figure<\/p>\n\n\n\n (d)<\/p>\n\n\n\n The reactance of a series LC circuit is given by,<\/p>\n\n\n\n X=XL-XC=\u03c9L-1\u03c9C\u21d2X=2\u03c0fL-12\u03c0fCThe correct relation between the reactance and f<\/em> is represented by the graph in option (d).<\/p>\n\n\n\n X=0, when<\/p>\n\n\n\n XL=XC.<\/p>\n\n\n\n Thus plot d represent the reactance of a series LC combination.<\/p>\n\n\n\n A series AC circuit has a resistance of 4 \u03a9 and a reactance of 3 \u03a9. The impedance of the circuit is<\/p>\n\n\n\n (a) 5 \u03a9 (a) 5 \u03a9<\/p>\n\n\n\n The impedance of the circuit is given by<\/p>\n\n\n\n Z=R2+X2R=4 \u03a9 & X=3 \u03a9\u21d2Z=42+32=5 \u03a9<\/p>\n\n\n\n Transformers are used<\/p>\n\n\n\n (a) in DC circuits only (b) in AC circuits only<\/p>\n\n\n\n When a DC supply is provided to a transformer, there will be no change in flux with time across the coils of the transformer. So, there will be no induced emf in the secondary coil due to changing current in the primary coil. Hence, the transformer cannot operate in DC because of the violation of its working principle.<\/p>\n\n\n\n An alternating current is given by i<\/em> = i<\/em>1<\/sub> cos \u03c9t<\/em> + i<\/em>2<\/sub> sin \u03c9t<\/em>. The rms current is given by<\/p>\n\n\n\n (a)<\/p>\n\n\n\n i1+i22(b)<\/p>\n\n\n\n i1+i22(c)<\/p>\n\n\n\n i12+i222(d)<\/p>\n\n\n\n i12+i222<\/p>\n\n\n\n (c)<\/p>\n\n\n\n i12+i222Given: irms=\u222b0Ti2dt\u222b0Tdti=i1cos \u03c9t +i2 sin \u03c9tirms=\u222b0Ti1cos \u03c9t +i2 sin \u03c9t2dt\u222b0Tdtirms=\u222b0Ti12cos2\u03c9t +i22 sin2\u03c9t +2i1i2sin \u03c9t cos \u03c9t dt\u222b0Tdtirms=\u222b0Ti12(cos 2\u03c9t+1)2 +i22 (1-cos 2\u03c9t) 2 +i1i2sin 2\u03c9t dt\u222b0Tdt [\u2235cos2\u03c9t =(cos 2\u03c9t+1)2 , sin2\u03c9t=(1-cos 2\u03c9t) 2 ]We know that, T<\/em> = 2\u03c0<\/p>\n\n\n\n Integrating the above expression<\/p>\n\n\n\n irms=12i12 \u222b02\u03c01dt+\u222b02\u03c0cos 2\u03c9t dt + i22 \u222b02\u03c01 dt -\u222b02\u03c0 cos 2\u03c9t dt +i1i2\u222b02\u03c0 sin 2\u03c9t dt\u222b02\u03c0dtThe following integrals become zero<\/p>\n\n\n\n \u222b02\u03c0cos 2\u03c9t dt = 0=\u222b02\u03c0 sin 2\u03c9t dtTherefore, it becomes<\/p>\n\n\n\n irms=i122\u222b02\u03c01dt+i222\u222b02\u03c01dt\u222b02\u03c0dtirms=i122\u00d72\u03c0 +i222\u00d72\u03c02\u03c0\u21d2irms =i12+i222<\/p>\n\n\n\n An alternating current of peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i<\/em> can be used, where i<\/em> is<\/p>\n\n\n\n (a) 14 A (d) about 10 A<\/p>\n\n\n\n The rms value of an alternating current is equivalent to the constant current. So, the heating effect produced is actually measured in terms of the rms value, in case of alternating current. The constant current is, thus, equal to the rms value of alternating current, which is given by,<\/p>\n\n\n\n Irms=Ipeak2=142=9.9\u224810 A<\/p>\n\n\n\n A constant current of 2.8 A exists in a resistor. The rms current is<\/p>\n\n\n\n (a) 2.8 A (a) 2.8 A<\/p>\n\n\n\n The constant current is equal to the rms value of current. So, An inductor, a resistance and a capacitor are joined in series with an AC source. As the frequency of the source is slightly increased from a very low value, the reactance<\/p>\n\n\n\n (a) of the inductor increases (a) of the inductor increases<\/p>\n\n\n\n The reactance of an inductor is given by,<\/p>\n\n\n\n XL=\u03c9LAnd the reactance of a capacitor is given by,<\/p>\n\n\n\n XC=1\u03c9CHere, \u03c9<\/em> = 2\u03c0f<\/em> , where f<\/em> is the frequency of the source. So, when f<\/em> increases, \u03c9<\/em> increases. The reactance of a circuit is zero. It is possible that the circuit contains<\/p>\n\n\n\n (a) an inductor and a capacitor (a) an inductor and a capacitor (a) The reactance of a circuit containing a capacitor and an inductor is given by,<\/p>\n\n\n\n X=XL-XC=\u03c9L-1\u03c9CIf<\/p>\n\n\n\n \u03c9L=1\u03c9C, X<\/em> = 0. So, the circuit contains an inductor and a capacitor.<\/p>\n\n\n\n (b) For a circuit without a capacitor, reactance is given by,<\/p>\n\n\n\n X\u2019=XL=\u03c9L, which cannot be zero for an AC source.<\/p>\n\n\n\n (c) Similarly, for a circuit without an inductor, reactance is given by,<\/p>\n\n\n\n X\u201d=XC=1\u03c9C, which also cannot be zero.<\/p>\n\n\n\n (d) For a circuit without any capacitor and inductor, reactance, X\u2018<\/sup><\/em> = 0 ( L<\/em> = C <\/em>= 0)<\/p>\n\n\n\n In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a<\/p>\n\n\n\n (a) pure inductor (a) pure inductor For a pure inductive circuit, voltage leads the current by<\/p>\n\n\n\n 90\u00b0. So, the instantaneous current is zero when the instantaneous voltage is maximum. 90\u00b0.<\/p>\n\n\n\n Also, in a circuit containing a combination of inductor and capacitor, the current may lead or lag the voltage by<\/p>\n\n\n\n 90\u00b0, depending upon whether the voltage across the inductor or the capacitor is greater. Here too, there is a phase difference of<\/p>\n\n\n\n 90\u00b0between the voltage and current. Hence, the the instantaneous current is zero when the instantaneous voltage is maximum.<\/p>\n\n\n\n An inductor coil of some resistance is connected to an AC source. Which of the following quantities have zero average value over a cycle?<\/p>\n\n\n\n (a) Current (a) Current For a series L<\/em>\u2013R<\/em> circuit, the AC current can be given by,<\/p>\n\n\n\n i=i0sin\u03c9t From the graph, we can see that the average value of current over a cycle is zero.<\/p>\n\n\n\n Since it a series L<\/em>\u2013R<\/em> circuit, the phase difference between current and voltage is<\/p>\n\n\n\n \u03c02. The AC voltage can be given by,<\/p>\n\n\n\n V=V0cos\u03c9t From the graph, we can see that the average value of voltage over a cycle is also zero.<\/p>\n\n\n\n Joule\u2019s heat through the resistor is given by,<\/p>\n\n\n\n Havg=irms2R, which is non zero.<\/p>\n\n\n\n Similarly, magnetic energy stored in the inductor is given by,<\/p>\n\n\n\n Uavg=12Lirms2, which is also non-zero.<\/p>\n\n\n\n The AC voltage across a resistance can be measured using<\/p>\n\n\n\n (a) a potentiometer (b) a hot-wire voltmeter<\/p>\n\n\n\n Only a hot-wire voltmeter can be used to measure an AC voltage across a resistor.<\/p>\n\n\n\n To convert mechanical energy into electrical energy, one can use<\/p>\n\n\n\n (a) DC dynamo (a) DC dynamo An AC or DC dynamo can be used to convert mechanical energy to electrical energy. An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. The average power delivered by the source<\/p>\n\n\n\n (a) must be 1000 W (b) may be 1000 W The average power delivered by an AC source is given by,<\/p>\n\n\n\n Pavg=VrmsIrmscos\u03d5Given: \u21d2Pavg=1000cos\u03d5<\/p>\n\n\n\n \u2235\u03d5\u2208-\u03c02,\u03c02<\/p>\n\n\n\n \u21d2cos\u03d5\u22080,1<\/p>\n\n\n\n \u22340\u2264Pavg\u22641000<\/p>\n\n\n\n Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.<\/p>\n\n\n\n Frequency of alternating current, f<\/em> = 50 Hz iis given by, irmsis given by,<\/p>\n\n\n\n irms=i02 \u20261 i02=i0sin\u03c9t\u21d212 = sin\u03c9t = sin\u03c04\u21d2\u03c04 = \u03c9t\u21d2t =\u03c04\u03c9 = \u03c04\u00d72\u03c0f \u2235\u03c9=2\u03c0f = 18f = 18\u00d750 =1400=0.0025 s =2.5 ms<\/p>\n\n\n\n The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.<\/p>\n\n\n\n RMS value of voltage, E<\/em>rms<\/sub> = 220 V, E0is given by, E0=Erms 2\u21d2 E0=2\u00d7220\u21d2 E0=311.08 V=311 V(b) Voltage<\/p>\n\n\n\n Eis given by, As E=E02, E02 = E0sin \u03c9t\u21d2\u03c9t = \u03c04\u21d2t=\u03c04\u03c9=\u03c04\u00d72\u03c0f\u21d2 t=\u03c08\u03c050=1400\u21d2 t= 2.5 msThus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.<\/p>\n\n\n\n A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.<\/p>\n\n\n\n Power of the bulb, P<\/em> = 60 W \u2234 R=V2P=220\u00d722060 =806.67Peak value of voltage<\/p>\n\n\n\n E0is given by,<\/p>\n\n\n\n E0=Erms2 =<\/p>\n\n\n\n 220\u00d72 = 311.08 i0is,<\/p>\n\n\n\n i0=E0R<\/p>\n\n\n\n \u21d2i0=311.08806.67=0.39 A<\/p>\n\n\n\n An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?<\/p>\n\n\n\n Voltage across the electric bulb, E<\/em> = 12 volts ithrough a resistor is equal to heat produced by passing a constant current<\/p>\n\n\n\n irmsthrough the same resistor. If R<\/em> is the resistance of the electric bulb and T<\/em> is the temperature, then<\/p>\n\n\n\n i2RT=irms2RT\u21d2E2R2=Erms2R2\u21d2E2=E022 \u2235E2rms=E202\u21d2E02=2E2\u21d2E02=2\u00d7122=2\u00d7144\u21d2E0=2\u00d7144 =16.97=17 VThus , peak value of voltage is 17 V.<\/p>\n\n\n\n The peak power consumed by a resistive coil, when connected to an AC source, is 80 W. Find the energy consumed by the coil in 100 seconds, which is many times larger than the time period of the source.<\/p>\n\n\n\n Peak power of the resistive coil,<\/p>\n\n\n\n P0 =80 WTime, t<\/em> = 100 s Prmsis given by,<\/p>\n\n\n\n Prms = P02, \u2234 Prms=P02=40 WEnergy consumed<\/p>\n\n\n\n Eis given by, The dielectric strength of air is 3.0 \u00d7 106<\/sup> V\/m. A parallel-plate air-capacitor has area 20 cm2<\/sup> and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.<\/p>\n\n\n\n Given:Answer:<\/h4>\n\n\n\n
Current, i<\/em> = i<\/em>0<\/sub> sin \u03c9t<\/em> or i<\/em> = i<\/em>0<\/sub> cos (\u03c9t <\/em><\/p>\n\n\n\n
where Irms<\/sub><\/em> = rms value of current
Vrms<\/sub><\/em>= rms value of voltage
\u00cf\u2022<\/em> = phase difference between current and voltage
Here, \u00cf\u2022<\/em> =<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Let the inductance in the circuit be L<\/em>
Let the net impedence of the circuit be<\/p>\n\n\n\n
R<\/em> = resistance in the circuit
X<\/em>L<\/sub> = reactance due to inductor
X<\/em>C<\/sub> = reactance due to capacitor
The magnitude of the voltage across the inductor is given by<\/p>\n\n\n\n
Thus, the voltage across the inductor can be written as<\/p>\n\n\n\n
If<\/p>\n\n\n\n
Therefore if magnitude of<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
where C<\/em> = capacitance
K<\/em> = dielectric constant
A<\/em> = area of plates
d<\/em> = distance between the plates.<\/p>\n\n\n\n
Rms current,<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Page No 329:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 10:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/11216.png\")
<\/p>\n\n\n\nQuestion 12:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 13:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 14:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 1:<\/h4>\n\n\n\n
(b) AC
(c) DC as well as AC
(d) neither AC nor DC<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 2:<\/h4>\n\n\n\n
\u03b5 = \u03b50<\/sub> [cos (100 \u03c0 s\u22121<\/sup>)t<\/em> + cos (500 \u03c0 s\u22121<\/sup>)t<\/em>]
is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be i <\/em>= i<\/em>1<\/sub> cos [(100 \u03c0 s\u22121<\/sup>)t<\/em> + \u03c61<\/sub>) + i<\/em>2<\/sub> cos [(500\u03c0 s\u22121<\/sup>)t<\/em> + \u00cf\u20222<\/sub>]. So,<\/p>\n\n\n\n
(b) i<\/em>1<\/sub> = i<\/em>2<\/sub>
(c) i<\/em>1<\/sub> < i<\/em>2<\/sub>
(d) The information is insufficient to find the relation between i<\/em>1<\/sub> and i<\/em>2<\/sub>.<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
(b) about 160 V
(c) about 310 V
(d) 440 V<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Vrms<\/sub><\/em> = 220 V
The peak value of voltage is given by,<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(b) may be zero
(c) is never zero
(d)<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
The average voltage over the given time,<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/4578.png\")
![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/4_b4-1.png\")
From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(b) 50 Hz
(c) 200 Hz
(d) 500 Hz<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
![\"\"\/](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/5458.png\")
<\/figure>\n\n\n\n
So, in this case, T<\/em>\/4 = 5 ms<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Question 7:<\/h4>\n\n\n\n
(b) 7 \u03a9
(c) 12\/7 \u03a9
(d) 7\/12 \u03a9<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 8:<\/h4>\n\n\n\n
(b) in AC circuits only
(c) in both DC and AC circuits
(d) neither in DC nor in AC circuits<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 9:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
i<\/em> = i<\/em>1<\/sub> cos \u03c9t<\/em> + i<\/em>2<\/sub> sin \u03c9t<\/em>
The rms value of current is given by,<\/p>\n\n\n\nQuestion 10:<\/h4>\n\n\n\n
(b) about 20 A
(c) 7 A
(d) about 10 A<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 11:<\/h4>\n\n\n\n
(b) about 2 A
(c) 1.4 A
(d) undefined for a direct current<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Irms<\/sub><\/em> = 2.8 A<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
(b) of the resistor increases
(c) of the capacitor increases
(d) of the circuit increases<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
\u2234 XL<\/sub><\/em> will increase and XC<\/sub><\/em> will decrease.<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
(b) an inductor but no capacitor
(c) a capacitor but no inductor
(d) neither an inductor nor a capacitor<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) neither an inductor nor a capacitor<\/p>\n\n\n\nPage No 330:<\/h4>\n\n\n\n
Question 3:<\/h4>\n\n\n\n
(b) pure capacitor
(c) pure resistor
(d) combination of an inductor and a capacitor<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) pure capacitor
(d) combination of an inductor and a capacitor.<\/p>\n\n\n\n
Similar is the case with a purely capacitive circuit, in which, current leads the voltage by<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
(b) Induced emf in the inductor
(c) Joule heat
(d) Magnetic energy stored in the inductor<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) Induced emf in the inductor<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/4579.png\")
<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/4_b5.png\")
<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
(b) a hot-wire voltmeter
(c) a moving-coil galvanometer
(d) a moving-magnet galvanometer<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
Question 6:<\/h4>\n\n\n\n
(b) AC dynamo
(c) motor
(d) transformer<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(b) AC dynamo<\/p>\n\n\n\n
A motor converts electrical energy to mechanical energy and a transformer is used to step up or down the voltage or simply to transfer electrical energy.<\/p>\n\n\n\nQuestion 7:<\/h4>\n\n\n\n
(b) may be 1000 W
(c) may be greater than 1000 W
(d) may be less than 1000 W<\/p>\n\n\n\nAnswer:<\/h4>\n\n\n\n
(d) may be less than 1000 W<\/p>\n\n\n\n
Vrms<\/sub> = 100 V
Irms<\/sub> = 10 A<\/p>\n\n\n\nQuestion 1:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Alternation current<\/p>\n\n\n\n
i<\/em> = i<\/em>0<\/sub>sin\u03c9t \u2026(1)
Here, i<\/em>0<\/sub> = peak value of current
Root mean square value of current<\/p>\n\n\n\n
On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:<\/p>\n\n\n\nQuestion 2:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Frequency of alternating current, f <\/em>= 50 Hz
(a) Peak value of voltage<\/p>\n\n\n\n
E0=Erms 2,
where E<\/em>rms<\/sub> = root mean square value of voltage<\/p>\n\n\n\n
E=E0sin\u03c9t,
where E<\/em>0<\/sub> = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.<\/p>\n\n\n\nQuestion 3:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, E<\/em>rms<\/sub> = 220 V
P = V<\/em>2<\/sup>R<\/em>,<\/em>
where R<\/em> = resistance of the bulb<\/p>\n\n\n\n
Now, maximum current through the filament<\/p>\n\n\n\nQuestion 4:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
Let E<\/em>0<\/sub> be the peak value of voltage.
We know that heat produced by passing an alternating current<\/p>\n\n\n\nQuestion 5:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n
RMS value of power<\/p>\n\n\n\n
where P<\/em>0<\/sub> = Peak value of power<\/p>\n\n\n\n
E<\/em> = P<\/em>rms<\/sub> \u00d7 t<\/em>
= 40 \u00d7 100
= 4000 J = 4.0 kJ<\/p>\n\n\n\nQuestion 6:<\/h4>\n\n\n\n
Answer:<\/h4>\n\n\n\n