An inductor of reactance 100Ω$100\Omega$, a capacitor of reactance 50Ω$50\Omega$, and a resistor of resistance 50Ω$50\Omega$ are connected in series with an AC source of 10~V$10\mathrm{~V}$, 50~Hz$50\mathrm{~Hz}$. Average power dissipated by the circuit is ______ W. [cite: 185, 186]

Related Formula

Core Logic

Since the load is purely resistive, the current is in phase with the voltage: I(t) = I₀ (100π t)$I(t) = I_0 \sin(100\pi t)$. We need the time difference between the instant current reaches (I₀)/(2)$\frac{I_0}{2}$ and the instant it reaches I₀$I_0$.

Step 1: Time for Half Peak

Step 2: Time for Peak

Step 3: Calculate Time Interval

The required time interval is Δ t = t₂ - t₁$\Delta t = t_2 - t_1$:

Given ω = 100π$\omega = 100\pi$:

Pattern Recognition

In an AC sine wave, going from 0 to peak takes T/4$T/4$. Going from 0 to half peak takes T/12$T/12$. Therefore, going from half peak to peak takes T/4 - T/12 = T/6$T/4 - T/12 = T/6$. Calculate T/6$T/6$ directly.

Chapter Mix

Class 12 Physics: Alternating Current

Related Formula

Core Logic

First, evaluate the secondary voltage induced by the transformer using the turns ratio. Then, apply basic DC voltage divider (or Ohm's law) logic to the secondary circuit to find the voltage drop V₀$V_0$ across the specific tapped resistance.

Step 1: Calculate Secondary Voltage

Step 2: Output Voltage Calculation

The secondary circuit contains two resistors in series (e.g., 15 ~kΩ$15 \mathrm{~k\Omega}$ and 7 ~kΩ$7 \mathrm{~k\Omega}$ forming 22 ~kΩ$22 \mathrm{~k\Omega}$ total, based on standard circuit values extracted from problem context). Current in secondary:

Output voltage V₀$V_0$ across the 7 ~kΩ$7 \mathrm{~k\Omega}$ resistor is:

Pattern Recognition

Two-step AC circuits: Step 1 transforms voltage perfectly (assume ideal transformer unless stated). Step 2 uses standard resistor scaling.

Chapter Mix

Class 12 Physics: Alternating Current Class 12 Physics: Current Electricity

Related Formula

Core Logic

First, establish the relationship between resistance R$R$ and initial inductive reactance XL$X_L$ using the first power factor. Then, double the frequency based on the new emf equation and recalculate the power factor.

Step 1: Initial Circuit State

Initial E = 25 (1000t)$E = 25\sin(1000t)$, giving ω₁ = 1000 ~rad/s$\omega_1 = 1000 \mathrm{~rad/s}$. Initial power factor θ = 1√(2) ⇒ θ = 45^°$\cos \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ$.

So, R = ω₁ L$R = \omega_1 L$.

Step 2: Second Circuit State

New E = 20 (2000t)$E = 20\sin(2000t)$, giving ω₂ = 2000 ~rad/s = 2ω₁$\omega_2 = 2000 \mathrm{~rad/s} = 2\omega_1$. New inductive reactance:

Calculate new power factor:

From trigonometric identity, θ' = 1√(1 + ² θ')$\cos \theta' = \frac{1}{\sqrt{1 + \tan^2 \theta'}}$:

Pattern Recognition

Power factor is fundamentally locked to the impedance triangle. If frequency doubles, XL$X_L$ doubles. The base leg R$R$ is constant, immediately stretching the triangle height and reducing the cosine.

Chapter Mix

Class 12 Physics: Alternating Current

Related Formula

where φ$\phi$ is the phase difference between voltage and current.

Core Logic

From the given equations: V₀ = 20 V$V_0 = 20 \text{ V}$ I₀ = 10 A$I_0 = 10 \text{ A}$ φ = (π)/(3) = 60°$\phi = \frac{\pi}{3} = 60^{\circ}$

Step 1: Calculate Power

Pattern Recognition

Average power in AC is half the product of peak voltage and peak current, scaled by the power factor (φ$\cos \phi$). Memorize P = (1)/(2) V₀ I₀ φ$\langle P \rangle = \frac{1}{2} V_0 I_0 \cos\phi$ to bypass RMS fractional clutter.

Chapter Mix

Class 12 Physics: Alternating Current