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In an ac circuit emf and current are E = 5 cos wt volt and I = 2 sin wt ampere respectively. The average power dissipated in this circuit will be
- a)10 W
- b)2.5 W
- c)5 W
- d)Zero
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In an ac circuit emf and current are E = 5 cos wt volt and I = 2 sin w...
The voltage can be written as,
V=5(sinωt+(π/2))
The angle between the voltage and the current is 90O.
The power dissipated in the circuit is given as,
P=VrmsIrmscosϕ
=VrmsIrmscos90O
=0
Thus, the power dissipated in the circuit is 0.
V=5(sinωt+(π/2))
The angle between the voltage and the current is 90O.
The power dissipated in the circuit is given as,
P=VrmsIrmscosϕ
=VrmsIrmscos90O
=0
Thus, the power dissipated in the circuit is 0.
Most Upvoted Answer
In an ac circuit emf and current are E = 5 cos wt volt and I = 2 sin w...
Given,
Emf E = 5 cos wt volts
Current I = 2 sin wt amperes
To find: Average power dissipated in the circuit
Formula used: The average power P_avg in an AC circuit is given by P_avg = E_eff x I_eff x cos ϕ, where E_eff and I_eff are the effective or RMS values of voltage and current respectively and ϕ is the phase difference between them.
1. Calculating the effective values of voltage and current
The effective or RMS value of a sinusoidal waveform is given by the formula: V_eff = V_m/√2, where V_m is the maximum value of the voltage or current.
Here,
Maximum value of voltage V_m = 5 volts
Maximum value of current I_m = 2 amperes
Therefore,
Effective value of voltage E_eff = V_m/√2 = 5/√2 volts
Effective value of current I_eff = I_m/√2 = 2/√2 amperes
2. Calculating the phase difference between voltage and current
The phase difference ϕ between voltage and current can be calculated using the formula: ϕ = θ_v - θ_i, where θ_v and θ_i are the phase angles of voltage and current respectively.
Here,
θ_v = 0 degrees (since voltage waveform is a cosine function)
θ_i = -90 degrees (since current waveform is a sine function)
Therefore,
ϕ = θ_v - θ_i = 0 - (-90) = 90 degrees
3. Calculating the average power dissipated
Using the formula P_avg = E_eff x I_eff x cos ϕ, we get
P_avg = (5/√2) x (2/√2) x cos 90 degrees
P_avg = 0 Watt (since cos 90 degrees = 0)
Hence, the average power dissipated in the circuit is zero.
Emf E = 5 cos wt volts
Current I = 2 sin wt amperes
To find: Average power dissipated in the circuit
Formula used: The average power P_avg in an AC circuit is given by P_avg = E_eff x I_eff x cos ϕ, where E_eff and I_eff are the effective or RMS values of voltage and current respectively and ϕ is the phase difference between them.
1. Calculating the effective values of voltage and current
The effective or RMS value of a sinusoidal waveform is given by the formula: V_eff = V_m/√2, where V_m is the maximum value of the voltage or current.
Here,
Maximum value of voltage V_m = 5 volts
Maximum value of current I_m = 2 amperes
Therefore,
Effective value of voltage E_eff = V_m/√2 = 5/√2 volts
Effective value of current I_eff = I_m/√2 = 2/√2 amperes
2. Calculating the phase difference between voltage and current
The phase difference ϕ between voltage and current can be calculated using the formula: ϕ = θ_v - θ_i, where θ_v and θ_i are the phase angles of voltage and current respectively.
Here,
θ_v = 0 degrees (since voltage waveform is a cosine function)
θ_i = -90 degrees (since current waveform is a sine function)
Therefore,
ϕ = θ_v - θ_i = 0 - (-90) = 90 degrees
3. Calculating the average power dissipated
Using the formula P_avg = E_eff x I_eff x cos ϕ, we get
P_avg = (5/√2) x (2/√2) x cos 90 degrees
P_avg = 0 Watt (since cos 90 degrees = 0)
Hence, the average power dissipated in the circuit is zero.
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Community Answer
In an ac circuit emf and current are E = 5 cos wt volt and I = 2 sin w...
Putting valueof cos phy =90
hence we got answer 0. for average pwr formula is irms×vrms×cosphi
hence we got answer 0. for average pwr formula is irms×vrms×cosphi
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