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The voltage across a load is v(t) = 60(ωt-10∘)cos V and the current through the load in the direction of the voltage drop is i(t)=1.5cos(ωt+50∘)A The reactive power is
- a)27.62 VAR leading
- b)38.97 VAR leading
- c)38.97 VAR lagging
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The voltage across a load is v(t) = 60(ωt-10)cos V and the curre...
The r.m.s. values of voltage and current phasors are :
Vr.m.s.=60/√2∠-10∘V;
Ir.m.s.=1.5√2∠-50∘
∴ Complex power,
S = Vr.m.s.I*r.m.s.
=60√2∠-10∘×1.5/√2∠-50∘
Complex power,
S=45∠-60∘VA-(22.5-j38.97)VA.
Therefore, reactive power = 38.97 VAR, leading because reactive power is negative.
Vr.m.s.=60/√2∠-10∘V;
Ir.m.s.=1.5√2∠-50∘
∴ Complex power,
S = Vr.m.s.I*r.m.s.
=60√2∠-10∘×1.5/√2∠-50∘
Complex power,
S=45∠-60∘VA-(22.5-j38.97)VA.
Therefore, reactive power = 38.97 VAR, leading because reactive power is negative.
Most Upvoted Answer
The voltage across a load is v(t) = 60(ωt-10)cos V and the curre...
1 - e^(-t/RC)) volts, where t is time in seconds, R is the resistance in ohms, and C is the capacitance in farads.
To find the time constant of the circuit, we can use the formula:
τ = RC
where τ is the time constant in seconds.
Since the voltage equation is given as v(t) = 60(1 - e^(-t/RC)), we can rearrange it to solve for τ:
v(t) = 60(1 - e^(-t/RC))
v(t)/60 = 1 - e^(-t/RC)
e^(-t/RC) = 1 - v(t)/60
Taking the natural logarithm of both sides:
-ln(1 - v(t)/60) = t/RC
Multiplying both sides by RC:
t = -RC ln(1 - v(t)/60)
Now we can plug in values for v(t) and solve for τ:
v(t) = 60(1 - e^(-t/RC))
When t = τ, e^(-t/RC) = 1/e, so:
v(τ) = 60(1 - 1/e)
Solving for τ:
τ = RC = t/(-ln(1 - v(τ)/60))
τ = t/(-ln(1 - 60(1 - 1/e)/60))
τ ≈ 0.693 RC
Therefore, the time constant of the circuit is approximately 0.693 times the product of the resistance and capacitance.
To find the time constant of the circuit, we can use the formula:
τ = RC
where τ is the time constant in seconds.
Since the voltage equation is given as v(t) = 60(1 - e^(-t/RC)), we can rearrange it to solve for τ:
v(t) = 60(1 - e^(-t/RC))
v(t)/60 = 1 - e^(-t/RC)
e^(-t/RC) = 1 - v(t)/60
Taking the natural logarithm of both sides:
-ln(1 - v(t)/60) = t/RC
Multiplying both sides by RC:
t = -RC ln(1 - v(t)/60)
Now we can plug in values for v(t) and solve for τ:
v(t) = 60(1 - e^(-t/RC))
When t = τ, e^(-t/RC) = 1/e, so:
v(τ) = 60(1 - 1/e)
Solving for τ:
τ = RC = t/(-ln(1 - v(τ)/60))
τ = t/(-ln(1 - 60(1 - 1/e)/60))
τ ≈ 0.693 RC
Therefore, the time constant of the circuit is approximately 0.693 times the product of the resistance and capacitance.
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The voltage across a load is v(t) = 60(ωt-10)cos V and the current through the load in the direction of the voltage drop isi(t)=1.5cos(ωt+50)A The reactive power isa)27.62 VAR leadingb)38.97 VAR leadingc)38.97 VAR laggingd)none of the aboveCorrect answer is option 'B'. Can you explain this answer?
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