Physics-
General
Easy

Question

If i subscript 1 end subscript equals 3 s i n invisible function application omega t and i subscript 2 end subscript equals 4 c o s invisible function application omega t, then i subscript 3 end subscript is

  1. 5 s i n invisible function application open parentheses omega t plus 53 to the power of ring operator end exponent close parentheses    
  2. 5 s i n invisible function application open parentheses omega t plus 37 to the power of ring operator end exponent close parentheses    
  3. 5 s i n invisible function application open parentheses omega t plus 45 to the power of ring operator end exponent close parentheses    
  4. 5 c o s invisible function application open parentheses omega t plus 53 to the power of ring operator end exponent close parentheses    

The correct answer is: 5 s i n invisible function application open parentheses omega t plus 53 to the power of ring operator end exponent close parentheses


    From Kirchoff’s current law
    table row cell i subscript 3 end subscript equals i subscript 1 end subscript plus i subscript 2 end subscript equals 3 s i n invisible function application omega t plus 4 s i n invisible function application open parentheses omega t plus 90 to the power of ring operator end exponent close parentheses end cell row cell equals square root of 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent plus 2 open parentheses 3 close parentheses open parentheses 4 close parentheses cos invisible function application 90 to the power of ring operator end exponent end root sin invisible function application open parentheses omega t plus ϕ close parentheses end cell row cell blank w h e r e blank t a n invisible function application ϕ equals fraction numerator 4 sin invisible function application 90 to the power of ring operator end exponent over denominator 3 plus 4 cos invisible function application 90 to the power of ring operator end exponent end fraction equals fraction numerator 4 over denominator 3 end fraction end cell end table
    therefore i subscript 3 end subscript equals 5 s i n invisible function application open parentheses omega t plus 53 to the power of ring operator end exponent close parentheses

    Related Questions to study

    General
    physics-

    At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0 cos(omegat) with l subscript 0 end subscript = 1A and omega = 500 rad/s starts flowing in it with the initial direction shown in the figure.

    text At  end text t equals fraction numerator 7 pi over denominator 6 omega end fraction text  , end text the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 mu F, R = 10 capital omega and the battery is ideal with emf of 50 V, identify the correct statement(s).

    At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0 cos(omegat) with l subscript 0 end subscript = 1A and omega = 500 rad/s starts flowing in it with the initial direction shown in the figure.

    text At  end text t equals fraction numerator 7 pi over denominator 6 omega end fraction text  , end text the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 mu F, R = 10 capital omega and the battery is ideal with emf of 50 V, identify the correct statement(s).

    physics-General
    General
    physics-

    In the given circuit, the AC source has omega = 100 rad/s. considering the inductor and capacitor to be ideal, the correct choice(S) is(are)

    In the given circuit, the AC source has omega = 100 rad/s. considering the inductor and capacitor to be ideal, the correct choice(S) is(are)

    physics-General
    General
    physics-

    In an L–R series circuit, a sinusoidal voltage V equals V subscript 0 end subscript s i n invisible function application omega t is applied. It is given that L = 35 mH, R = 11capital omega, Vrms = 220 V, omega divided by 2 pi equals 50 H z and pi = 22/7. Find the amplitude of current in the steady state and obtain the phase difference between the current and the voltage. Also plot the variation of current for one cycle on the given graph.

    An AC voltage source of variable angular frequency omega and fixed amplitude V connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When omega is increased :

    In an L–R series circuit, a sinusoidal voltage V equals V subscript 0 end subscript s i n invisible function application omega t is applied. It is given that L = 35 mH, R = 11capital omega, Vrms = 220 V, omega divided by 2 pi equals 50 H z and pi = 22/7. Find the amplitude of current in the steady state and obtain the phase difference between the current and the voltage. Also plot the variation of current for one cycle on the given graph.

    An AC voltage source of variable angular frequency omega and fixed amplitude V connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When omega is increased :

    physics-General
    parallel
    General
    physics-

    When an AC source of emf E equals E subscript 0 end subscript s i n invisible function application left parenthesis 100 t right parenthesis is connected across a circuit, the phase difference between the E and the current i in the circuit is observed to be pi divided by 4 , as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C series, find the relationship between the two elements.

    When an AC source of emf E equals E subscript 0 end subscript s i n invisible function application left parenthesis 100 t right parenthesis is connected across a circuit, the phase difference between the E and the current i in the circuit is observed to be pi divided by 4 , as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C series, find the relationship between the two elements.

    physics-General
    General
    physics-

    An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q subscript 0 end subscript and then connected to the L and R as shown below :

    If a student plots graphs of the square of maximum charge open parentheses Q subscript M a x end subscript superscript 2 end superscript close parentheses on the capacitor with time (T) for two different values L subscript 1 end subscript and L subscript 2 end subscript open parentheses L subscript 1 end subscript greater than L subscript 2 end subscript close parentheses of L then which of the following represents this graph correctly ? (Plots are shematic and not drawn to scale)

    An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q subscript 0 end subscript and then connected to the L and R as shown below :

    If a student plots graphs of the square of maximum charge open parentheses Q subscript M a x end subscript superscript 2 end superscript close parentheses on the capacitor with time (T) for two different values L subscript 1 end subscript and L subscript 2 end subscript open parentheses L subscript 1 end subscript greater than L subscript 2 end subscript close parentheses of L then which of the following represents this graph correctly ? (Plots are shematic and not drawn to scale)

    physics-General
    General
    physics-

    In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected. When switch 'S' is closed then there is a phase difference of fraction numerator pi over denominator 4 end fraction between current and applied voltage and voltage across resister is fraction numerator 100 over denominator square root of 2 end fraction V When switch is open current and applied voltage are in same phase. Neglecting resistance of connecting wire answer the following questions :

    Peak voltage of applied voltage sources is :

    In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected. When switch 'S' is closed then there is a phase difference of fraction numerator pi over denominator 4 end fraction between current and applied voltage and voltage across resister is fraction numerator 100 over denominator square root of 2 end fraction V When switch is open current and applied voltage are in same phase. Neglecting resistance of connecting wire answer the following questions :

    Peak voltage of applied voltage sources is :

    physics-General
    parallel
    General
    physics-

    A voltage source V = V subscript 0 end subscript sin (100 t) is connected to a black box in which there can be either one element out of L, C, R or any two of them connected in series

    At steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

    If AC source is removed, the circuit is shorted for some time so that capacitor is fully discharged and then a battery of constant EMF is connected across the black box, at t = 0. The current in the circuit will –

    A voltage source V = V subscript 0 end subscript sin (100 t) is connected to a black box in which there can be either one element out of L, C, R or any two of them connected in series

    At steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

    If AC source is removed, the circuit is shorted for some time so that capacitor is fully discharged and then a battery of constant EMF is connected across the black box, at t = 0. The current in the circuit will –

    physics-General
    General
    physics-

    A voltage source V = V subscript 0 end subscript sin (100 t) is connected to a black box in which there can be either one element out of L, C, R or any two of them connected in series

    At steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

    Values of the parameters of the elements, present in the black box are –

    A voltage source V = V subscript 0 end subscript sin (100 t) is connected to a black box in which there can be either one element out of L, C, R or any two of them connected in series

    At steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

    Values of the parameters of the elements, present in the black box are –

    physics-General
    General
    physics-

    A voltage source V = V subscript 0 end subscript sin (100 t) is connected to a black box in which there can be either one element out of L, C, R or any two of them connected in series

    At steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

    The element(s) present in black box is/are :

    A voltage source V = V subscript 0 end subscript sin (100 t) is connected to a black box in which there can be either one element out of L, C, R or any two of them connected in series

    At steady state. the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown

    The element(s) present in black box is/are :

    physics-General
    parallel
    General
    physics-

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    Thermal energy produced by the resistance R in time duration 1 mus, using the source at resonant condition, is

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    Thermal energy produced by the resistance R in time duration 1 mus, using the source at resonant condition, is

    physics-General
    General
    physics-

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    Average energy stored by the inductor L subscript 2 end subscript (Source is at resonance frequency) is equal to

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    Average energy stored by the inductor L subscript 2 end subscript (Source is at resonance frequency) is equal to

    physics-General
    General
    physics-

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    If the ac source G is of 100 V rating at resonant frequency of the circuit, then average power supplied by the source is –

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    If the ac source G is of 100 V rating at resonant frequency of the circuit, then average power supplied by the source is –

    physics-General
    parallel
    General
    physics-

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    To increase resonant frequency of the circuit, some of the changes in the circuit are carried out. Which change(s) would certainly result in the increase in resonant frequency ?

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    To increase resonant frequency of the circuit, some of the changes in the circuit are carried out. Which change(s) would certainly result in the increase in resonant frequency ?

    physics-General
    General
    physics-

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    Current drawn from the ac source will be maximum if its angular frequency is –

    An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.

    Current drawn from the ac source will be maximum if its angular frequency is –

    physics-General
    General
    physics-

    Statement-1 : In a series R,L,C circuit if V subscript R end subscript, V subscript L end subscript , and V subscript C end subscript denote rms voltage across R, L and C repectively and V subscript S end subscript is the rms voltage across the source, then V subscript S end subscript equals V subscript R end subscript plus V subscript L end subscript plus V subscript C end subscript.

    Statement-2 : In AC circuits, kirch off voltage law is correct at every instant of time.

    Statement-1 : In a series R,L,C circuit if V subscript R end subscript, V subscript L end subscript , and V subscript C end subscript denote rms voltage across R, L and C repectively and V subscript S end subscript is the rms voltage across the source, then V subscript S end subscript equals V subscript R end subscript plus V subscript L end subscript plus V subscript C end subscript.

    Statement-2 : In AC circuits, kirch off voltage law is correct at every instant of time.

    physics-General
    parallel

    card img

    With Turito Academy.

    card img

    With Turito Foundation.

    card img

    Get an Expert Advice From Turito.

    Turito Academy

    card img

    With Turito Academy.

    Test Prep

    card img

    With Turito Foundation.