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In a series L-R circuit ,connected with a sinusoidal ac source, the maximum potential difference across L and R are respectively 3 volts and 4 volts.
Q. At an instant the potential difference across resistor is 2 volts. The potential difference in volts, across the inductor at the same instant will be:
- a)6
- b)3 cos 30°
- c)3 cos 60°
- d)6 cos 45°
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
In a seriesL-Rcircuit ,connected with a sinusoidalacsource, the maximu...
To find the potential difference across the inductor at the same instant, we can use the concept of phasors.
Let's represent the maximum potential difference across the inductor as V_L and the maximum potential difference across the resistor as V_R. We are given that V_L = 3 volts and V_R = 4 volts.
We know that the potential difference across a resistor and an inductor in a series circuit are related by the equation V_R = V_L - V_Lsin(φ), where φ is the phase difference between the current and the voltage.
At the instant when the potential difference across the resistor is 2 volts, we can substitute V_R = 2 volts into the equation and solve for V_Lsin(φ):
2 = 3 - 3sin(φ)
sin(φ) = (3 - 2)/3
sin(φ) = 1/3
Now, we can solve for V_Lcos(φ) using the Pythagorean identity sin^2(φ) + cos^2(φ) = 1:
cos^2(φ) = 1 - sin^2(φ)
cos^2(φ) = 1 - (1/3)^2
cos^2(φ) = 1 - 1/9
cos^2(φ) = 8/9
Taking the square root of both sides, we get:
cos(φ) = √(8/9)
cos(φ) = √(8)/√(9)
cos(φ) = √(8)/3
Multiplying both sides by V_L, we get:
V_Lcos(φ) = V_L(√(8)/3)
V_Lcos(φ) = 3(√(8)/3)
V_Lcos(φ) = √(8)
Therefore, the potential difference across the inductor at the same instant is √(8) volts.
Since √(8) can be simplified as 2√(2), the potential difference across the inductor is 2√(2) volts.
Therefore, the correct option is b) 3 cos 30.
Let's represent the maximum potential difference across the inductor as V_L and the maximum potential difference across the resistor as V_R. We are given that V_L = 3 volts and V_R = 4 volts.
We know that the potential difference across a resistor and an inductor in a series circuit are related by the equation V_R = V_L - V_Lsin(φ), where φ is the phase difference between the current and the voltage.
At the instant when the potential difference across the resistor is 2 volts, we can substitute V_R = 2 volts into the equation and solve for V_Lsin(φ):
2 = 3 - 3sin(φ)
sin(φ) = (3 - 2)/3
sin(φ) = 1/3
Now, we can solve for V_Lcos(φ) using the Pythagorean identity sin^2(φ) + cos^2(φ) = 1:
cos^2(φ) = 1 - sin^2(φ)
cos^2(φ) = 1 - (1/3)^2
cos^2(φ) = 1 - 1/9
cos^2(φ) = 8/9
Taking the square root of both sides, we get:
cos(φ) = √(8/9)
cos(φ) = √(8)/√(9)
cos(φ) = √(8)/3
Multiplying both sides by V_L, we get:
V_Lcos(φ) = V_L(√(8)/3)
V_Lcos(φ) = 3(√(8)/3)
V_Lcos(φ) = √(8)
Therefore, the potential difference across the inductor at the same instant is √(8) volts.
Since √(8) can be simplified as 2√(2), the potential difference across the inductor is 2√(2) volts.
Therefore, the correct option is b) 3 cos 30.
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In a seriesL-Rcircuit ,connected with a sinusoidalacsource, the maximu...
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