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Alternating Current

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Physics Part I

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Physics

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Class 10 Class 12
The instantaneous current and voltage of an a.c. circuit are given by
i = 10 sin 314 t A and
v = 50 sin (314 t + π/2)V.
What is the power dissipation in the circuit?

Given,
Instantaneous current, i = 10 sin 314 t A

Instantaneous voltage, V = 50 sin (314 t + π/2) V 

Since, 
I = IO sin ωt and,V = VO sin (ωt +ϕ)
Therefore, from the above equations we get 

Peak current, peak voltage and ϕ as, 
i0 = 10,    V0 =50   and  ϕ = π2 

Power dissipation in the circuit is given by
Ev = E0 2 and  I0 = I o2 

Therefore, 

P = 502 102 cos π2   = 0          
                                                     cos π2 = 0

2248 Views

A 100 volt a.c. source of frequency 500 hertz is connected to LCR circuit with L = 8.1 millihenry, C = 12.5 micro farad and R = 10 ohm, all connected in series. Find the potential difference across the resistance.

Given, an LCR series circuit.
Rms value of voltage, V = 100 V
Frequency = 500 Hz

The impedance of LCR circuit is given by
               box enclose straight Z space equals space square root of open square brackets straight R squared plus open parentheses straight X subscript straight L minus straight X subscript straight C close parentheses squared close square brackets end root end enclose 

where     box enclose straight X subscript straight L space equals space ωL end enclose space equals space 2 πfL space space space space
                 equals space 2 space cross times space 3.14 space cross times space 500 space cross times space left parenthesis space 8.1 space cross times space 10 to the power of negative 3 end exponent right parenthesis
equals space 25.4 space ohm 

and,       box enclose straight X subscript straight C space equals space 1 over ωC end enclose space equals space fraction numerator 1 over denominator 2 πfC end fraction
                 equals space fraction numerator 1 over denominator 2 cross times 3.14 cross times 400 cross times left parenthesis 12.5 space cross times space 10 to the power of negative 6 end exponent right parenthesis end fraction
                  = 25.4 ohm 

therefore Impedence, straight Z space equals space square root of open square brackets left parenthesis 10 right parenthesis squared plus left parenthesis 25.4 minus 25.4 right parenthesis squared close square brackets end root space equals space 10 space ohm
Rms value of current, straight I subscript rms space equals space straight E subscript rms over straight Z space equals space fraction numerator 100 space volt over denominator 10 space ohm end fraction space equals space 10 space amp. 

Potential difference across resistance 

straight V subscript straight R space equals space straight I subscript rms space cross times space straight R space
space space space space space space equals space 10 space amp space cross times space 10 space ohm
space space space space space space equals space 100 space volt.
187 Views

A resistance of 10 ohm is joined in series with an inductance of 0.5 henry. What capacitance should be put in series with the combination to obtain the maximum current? What will be the potential difference across the resistance, inductance and capacitor? The current is being supplied by 200 volts and 50 cycles per second mains.

The current in the circuit would be maximum when XL = Xc

i.e.,                   ωL = 1ωC,  or   C = 1ω2L 
Therefore,

            
             C = 1(2πf)2L    = 1(2×3.14×50)2×0.5    = 20.24 × 10-6 farad. 

Here, ωL = 1ωC.
So the impedance Z of the circuit,

  Z = R2+ωL -1ωC2 and  R = 10 ohm 

I = ER= 20010 = 20 amp.     
              
Potential difference across resistance 

VR = I × R
    = 20 × 10
    = 200 volt  

Potential difference across inductance 

VL = ωL × I      = (2π × 50 × 0.5) × 20     = 3142 volt. 

Potential difference across condenser 

VC = 1ωC      = I × ωL      = 3142 volt.

140 Views

An inductor 200 μH, capacitor 500 μF, resistor 10 Ω are connected in series with a 100 V, variable frequency a.c. source. Calculate the
(i) frequency at which the power factor of the circuit is unity
(ii) current amplitude at this frequency
(iii) Q-factor

Given,
Inductor, L = 200 μH
Capacitor, C = 500 μF
Resistor, R = 10 Ω 
Effective voltage, V = 100 V

(i) Power factor,  cos θ = RZ = 1

So, R = Z 

 R = R2+ωL - 1ωC2 
   R2 = R2+ωL - 1ωC2 
  ωL = 1ωC 
  
  ω2 = 1LC   or  ω = 1LC 
 2πv = 1LC

      v = 12πLC

  ω0 = 12πLC      = 1200 × 10-6× 500 × 10-6      = 3.16 ×10-3 rad s-1 

 ω0 = 3.16 × 10-3 rad/s

(ii) The current amplitude at this frequency,  

I0 = VR = 10010 = 10 A 

(iii) The Q-factor, 

Q = XLR = ω0LR    = 3.16 × 10-3 × 200 × 10-610   = 6.32 × 10-8

351 Views

An LCR series circuit with 100 Ω resistance is connected to an a.c. source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the current and power dissipated in LCR circuit.


Given, an LCR series circuit.
Resistance, R = 100 Ω 
Rms voltage, V = 200 V 
Angular frequency = 300 radians per second.
Current lags behind the voltage by 60o 

Using the formula,

                 tan 60° = ωLR

or,             tan 60° = 1/ωCR
                    ωL = 1ωC 

Impedance of circuit, 

Z = R2+ωL -1ωC2 = R  

Current in the circuit,

I0 = V0Z = V0R =200100     = 2 amp. 

Average power, P = 12 V0 I0 cos ϕ 

But, 
tan ϕ = ωL -(1/ωC)R = 0   (cos ϕ = 1) 

Now,  P = 12×200×2×1 = 200 watt.

590 Views

A resistor of resistance R, an inductor inductance L and a capacitor of capacitance C all are connected in series with an a.c. supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find
(a) the potential difference across R, L and C
(b) the impedance of the circuit
(c) the voltage of a.c. supply
(d) phase angle


Given, an LCR circuit where all the components are connected in series with an a.c. supply.
Resistance, R = 16 Ω 
Inductive reactance, χL = 24 ohm
Capacitive reactance, χC = 12 Ω 
Current flowing in the circuit, I = 5 A

(a) Potential difference across resistance,
VR = iR
    = 5 × 16
    = 80 volt 

Potential difference across inductance,
VL = i × (ωL)
    = 5 × 24
    = 120 volt 

Potential difference across condenser,
VC = i × 1ωC      = 5×12       = 60 volt 

(b) The impedance of the circuit is given as 

Z = R2+ωL -1ωC2    = (16)2+(24-12)2 = 20 ohm 

(c) The voltage of a.c. supply is given by 

V = iz
   = 5 × 20
   = 100 volt 

(d) Phase angle 

ϕ = tan-1ωL -1ωCR
= tan-124-1216= tan-1(0.75) = 36°46'.

161 Views