In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t `i = 20 sin( 30t-(pi)/(4))` In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :
A
50,0
B
50,10
C
`(1000)/(sqrt(2)),10`
D
`(50)/(sqrt(2)),0`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the given problem, we will follow these steps:
### Step 1: Identify the given parameters
The instantaneous EMF and current are given as:
- \( e(t) = 100 \sin(30t) \)
- \( i(t) = 20 \sin(30t - \frac{\pi}{4}) \)
From these equations, we can identify:
- The peak voltage \( E_0 = 100 \) V
- The peak current \( I_0 = 20 \) A
### Step 2: Calculate the phase difference \( \phi \)
The phase difference \( \phi \) between the voltage and current can be determined from the current equation:
- \( i(t) = 20 \sin(30t - \frac{\pi}{4}) \)
Thus, the phase angle \( \phi = \frac{\pi}{4} \) radians.
### Step 3: Calculate the average power consumed in the circuit
The formula for average power \( P \) in an AC circuit is given by:
\[
P = \frac{E_0 I_0}{2} \cos(\phi)
\]
Substituting the values we have:
- \( E_0 = 100 \)
- \( I_0 = 20 \)
- \( \phi = \frac{\pi}{4} \) (where \( \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \))
Now substituting these values into the formula:
\[
P = \frac{100 \times 20}{2} \cdot \cos\left(\frac{\pi}{4}\right) = \frac{2000}{2} \cdot \frac{1}{\sqrt{2}} = 1000 \cdot \frac{1}{\sqrt{2}} = \frac{1000}{\sqrt{2}} \text{ W}
\]
### Step 4: Calculate the wattless current
The wattless current \( I_w \) can be calculated using the formula:
\[
I_w = I_{\text{rms}} \cdot \sin(\phi)
\]
Where \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \).
Calculating \( I_{\text{rms}} \):
\[
I_{\text{rms}} = \frac{20}{\sqrt{2}} \text{ A}
\]
Now substituting into the wattless current formula:
\[
I_w = \frac{20}{\sqrt{2}} \cdot \sin\left(\frac{\pi}{4}\right) = \frac{20}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{20}{2} = 10 \text{ A}
\]
### Final Answers
- Average Power: \( \frac{1000}{\sqrt{2}} \) W
- Wattless Current: \( 10 \) A
Topper's Solved these Questions
JEE MAINS
JEE MAINS PREVIOUS YEAR ENGLISH|Exercise Chemistry|1 Videos
JEE MAIN
JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|452 Videos
JEE MAINS 2020
JEE MAINS PREVIOUS YEAR ENGLISH|Exercise PHYSICS|250 Videos
Similar Questions
Explore conceptually related problems
In an AC circuit, the instantaneous values of e.m.f and current are e=200sin314t volt and i=sin(314t+(pi)/3) ampere. The average power consumed in watt is
In an AC circuit, the instantaneous values of e.m.f and current are e=200sin314t volt and i=sin(314t+(pi)/3) ampere. The average power consumed in watt is
In an AC circuit the instantaneous values of emf and current are e=200sin300t volt and i=2sin(300t+(pi)/(3)) amp The average power consumed (in watts) is
In an a.c. circuit the e.m.f. € and the current (i) al any instant are-given respectively by e = E_(0)sin omegat I = l_(0)sin (omegat-phi) The average power in the circuit over one cycle of a.c. is
In an a.c circuit, V & I are given by V = 100 sin (100 t) volt. I = 100 sin (100 t + (pi)/(3)) mA The power dissipated in the circuit is:
The instantaneous current and volatage of an AC circuit are given by i = 10 sin (314 t) A and V = 100 sin (314 t) V What is the power dissipation in the circuit?
The instantaneous current and voltage of an a.c. circuit are given by i=10 sin 3000 t A and v=200 sin 300 t V. What is the power dissipation in the circuit?
In an AC circuit V and I are given by V = 100 sin 100t V and I = 100 sin (100t + pi//3) mA. The power dissipated in the circuit is
In a circuit, the instantaneous values of alternating current and voltages in a circuit is given by I = (1)/(sqrt2) sin (100 pi t) A and E = (1)/(sqrt2) sin (100 pi t + (pi)/(3)) V . The average power in watts consumed in the circui is
In an AC circuit, the applied potential difference and the current flowing are given by V=20sin100t volt , I=5sin(100t-pi/2) amp The power consumption is equal to
JEE MAINS PREVIOUS YEAR ENGLISH-JEE MAINS-Chemistry