Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Supply voltage, V = 110 V Frequency, ν = 60 Hz Angular frequency, ω= 2πν Capacitive reactance, XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω rms value of current is given as : I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) ) = 110 x (2π x 60 x 60 x 10-6)=Read more
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω= 2πν Capacitive reactance,
XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω
rms value of current is given as :
I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) )
= 110 x (2π x 60 x 60 x 10-6)= 2.49 A Hence, the rms value of current is 2.49 A.
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H Supply voltage, V = 220 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω rms value of current is given as: I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A Hence ,the rms value of current in theRead more
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H
Supply voltage, V = 220 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν
Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω
rms value of current is given as:
I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A
Hence ,the rms value of current in the circuit is 15.92 A
Ans (a). Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as: V = V0/√ 2 = 300/√ 2 = 212.1 V Ans (b). The rms value of current is given as: I = 10 A Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
Ans (a).
Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as:
V = V0/√ 2 = 300/√ 2 = 212.1 V
Ans (b).
The rms value of current is given as: I = 10 A
Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
Resistance of the resistor, R = 100 Ω, Supply voltage, V = 220 V, Frequency ,ν = 50 Hz Ans (a). The rms value of current in the circuit is given as, I =V/R =220/100 = 2.20 A Ans (b). The net power consumed over a full cycle is given as : P = VI = 220 x 2.2 = 484 W
Resistance of the resistor, R = 100 Ω,
Supply voltage, V = 220 V,
Frequency ,ν = 50 Hz
Ans (a).
The rms value of current in the circuit is given as,
I =V/R =220/100 = 2.20 A
Ans (b).
The net power consumed over a full cycle is given as :
Line charge per unit length = λ = Total Charge/Length = Q/2πr Where, r = Distance of the point within the wheel Mass of the wheel = M Radius of the wheel = R Magnetic field, B = -B0 k At distance r, the magnetic force is balanced by the centripetal force i.e., BQv = Mv²/r Where, v = linear velocityRead more
Line charge per unit length = λ = Total Charge/Length = Q/2πr
Where, r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, B = -B0 k
At distance r, the magnetic force is balanced by the centripetal force i.e.,
A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F Supply voltage, V = 110 V Frequency, ν = 60 Hz Angular frequency, ω= 2πν Capacitive reactance, XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω rms value of current is given as : I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) ) = 110 x (2π x 60 x 60 x 10-6)=Read more
Capacitance of capacitor, C = 60 μF = 60 x 10-6 F
Supply voltage, V = 110 V
Frequency, ν = 60 Hz
Angular frequency, ω= 2πν
Capacitive reactance,
XC = 1/ωC = 1/2πνC =1/ (2π x 60 x 60 x 10-6) Ω
rms value of current is given as :
I = V/XC = 110/(1/ (2π x 60 x 60 x 10-6) )
= 110 x (2π x 60 x 60 x 10-6)= 2.49 A
See lessHence, the rms value of current is 2.49 A.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H Supply voltage, V = 220 V, Frequency, ν = 50 Hz, Angular frequency, ω = 2πν Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω rms value of current is given as: I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A Hence ,the rms value of current in theRead more
Inductance of inductor, L = 44 mH = 44 x 10⁻3 H
Supply voltage, V = 220 V,
Frequency, ν = 50 Hz,
Angular frequency, ω = 2πν
Inductive reactance, XL= ωL = 2πνL = 2π x 50 x 44 x 10-3 Ω
rms value of current is given as:
I = V/XL =220/(2π x 50 x 44 x 10-3) =15.92 A
Hence ,the rms value of current in the circuit is 15.92 A
See less(a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Ans (a). Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as: V = V0/√ 2 = 300/√ 2 = 212.1 V Ans (b). The rms value of current is given as: I = 10 A Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
Ans (a).
Peak voltage of the ac supply, Vo = 300 Vrms voltage is given as:
V = V0/√ 2 = 300/√ 2 = 212.1 V
Ans (b).
The rms value of current is given as: I = 10 A
Now, peak current is given as: I0 = √ 2 I = √ 2 x 10 = 14.1 A
See lessA 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?
Resistance of the resistor, R = 100 Ω, Supply voltage, V = 220 V, Frequency ,ν = 50 Hz Ans (a). The rms value of current in the circuit is given as, I =V/R =220/100 = 2.20 A Ans (b). The net power consumed over a full cycle is given as : P = VI = 220 x 2.2 = 484 W
Resistance of the resistor, R = 100 Ω,
Supply voltage, V = 220 V,
Frequency ,ν = 50 Hz
Ans (a).
The rms value of current in the circuit is given as,
I =V/R =220/100 = 2.20 A
Ans (b).
The net power consumed over a full cycle is given as :
P = VI = 220 x 2.2 = 484 W
See lessA line charge A per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22]. A uniform magnetic field extends over a circular region within the rim. It is given by, B = – Bo k (r ≤ a; a < R) = 0 (otherwise].What is the angular velocity of the wheel after the field is suddenly switched off?
Line charge per unit length = λ = Total Charge/Length = Q/2πr Where, r = Distance of the point within the wheel Mass of the wheel = M Radius of the wheel = R Magnetic field, B = -B0 k At distance r, the magnetic force is balanced by the centripetal force i.e., BQv = Mv²/r Where, v = linear velocityRead more
Line charge per unit length = λ = Total Charge/Length = Q/2πr
Where, r = Distance of the point within the wheel
Mass of the wheel = M
Radius of the wheel = R
Magnetic field, B = -B0 k
At distance r, the magnetic force is balanced by the centripetal force i.e.,
BQv = Mv²/r
Where,
v = linear velocity of the wheel
Therefore, B2πrλ = Mv/r
v =B2πλr²/M
Therefore ,Angular velocity , ω =v/R = B2πλr²/MR
For r ≤ a ≤ R, we get
ω = -(2πB0a²λ/MR) k^
See less