An inductor $20\, mH$, a capacitor $100\, \mu F$ and a resistor $50 \, \Omega$ are connected in series across a source of emf, $V\, = \,10 \,sin\, 314\, t$. The power loss in the circuit is

Question

Which of the following correctly represents the AND logic gate?

Answer 1

To determine the power loss in the given RLC circuit with an inductor $L = 20 \, \text{mH}$, a capacitor $C = 100 \, \mu \text{F}$, and a resistor $R = 50 \, \Omega$, connected to a source of emf $V = 10 \sin 314 t$, let's proceed step-by-step:

Calculate the angular frequency ($ \omega $):
The given voltage source is in the form $ V = 10 \sin 314 t $. Comparing it with the standard form $ V = V_0 \sin \omega t $, we identify $\omega = 314 \, \text{rad/s}$.
Calculate the reactances:
- Inductive Reactance ($X_L$): X_L = \omega L = 314 \times 20 \times 10^{-3} \, \Omega = 6.28 \, \Omega
- Capacitive Reactance ($X_C$): X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} \, \Omega = 31.85 \, \Omega
Calculate the impedance ($Z$) of the circuit:
The total impedance $Z$ of a series RLC circuit is given by:

Z = \sqrt{R^2 + (X_L - X_C)^2}

Substitute the values:

Z = \sqrt{50^2 + (6.28 - 31.85)^2} \, \Omega = \sqrt{50^2 + (-25.57)^2} \, \Omega = \sqrt{2500 + 653.62} \, \Omega = \sqrt{3153.62} \, \Omega \approx 56.15 \, \Omega
Calculate the current ($I$) in the circuit:
The maximum current $I$ is obtained from the maximum voltage $V_0 = 10 \, \text{V}$:

I = \frac{V_0}{Z} = \frac{10}{56.15} \approx 0.178 \, \text{A}
Determine the power loss ($P$) in the resistor:
The average power dissipated in the resistor is given by:

P = I^2 R = (0.178)^2 \times 50 \approx 0.79 \, \text{W}

Thus, the power loss in the circuit is 0.79 W, which matches option 2. Hence, the correct answer is 0.79 W.

Answer 2

To determine the power loss in the given RLC circuit with an inductor $L = 20 \, \text{mH}$, a capacitor $C = 100 \, \mu \text{F}$, and a resistor $R = 50 \, \Omega$, connected to a source of emf $V = 10 \sin 314 t$, let's proceed step-by-step:

Calculate the angular frequency ($ \omega $):
The given voltage source is in the form $ V = 10 \sin 314 t $. Comparing it with the standard form $ V = V_0 \sin \omega t $, we identify $\omega = 314 \, \text{rad/s}$.
Calculate the reactances:
- Inductive Reactance ($X_L$): X_L = \omega L = 314 \times 20 \times 10^{-3} \, \Omega = 6.28 \, \Omega
- Capacitive Reactance ($X_C$): X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} \, \Omega = 31.85 \, \Omega
Calculate the impedance ($Z$) of the circuit:
The total impedance $Z$ of a series RLC circuit is given by:

Z = \sqrt{R^2 + (X_L - X_C)^2}

Substitute the values:

Z = \sqrt{50^2 + (6.28 - 31.85)^2} \, \Omega = \sqrt{50^2 + (-25.57)^2} \, \Omega = \sqrt{2500 + 653.62} \, \Omega = \sqrt{3153.62} \, \Omega \approx 56.15 \, \Omega
Calculate the current ($I$) in the circuit:
The maximum current $I$ is obtained from the maximum voltage $V_0 = 10 \, \text{V}$:

I = \frac{V_0}{Z} = \frac{10}{56.15} \approx 0.178 \, \text{A}
Determine the power loss ($P$) in the resistor:
The average power dissipated in the resistor is given by:

P = I^2 R = (0.178)^2 \times 50 \approx 0.79 \, \text{W}

Thus, the power loss in the circuit is 0.79 W, which matches option 2. Hence, the correct answer is 0.79 W.

An inductor $20\, mH$, a capacitor $100\, \mu F$ and a resistor $50 \, \Omega$ are connected in series across a source of emf, $V\, = \,10 \,sin\, 314\, t$. The power loss in the circuit is

The Correct Option is B

Solution and Explanation

Top Questions on Electromagnetic induction

Questions Asked in NEET (UG) exam