A series L,R circuit connected with an ac source E = (25 1000t)V$E = (25\sin 1000t)\mathrm{V}$ has a power factor of 1√(2)$\frac{1}{\sqrt{2}}$. If the source of emf is changed to E = (20 2000t)V$E = (20\sin 2000t)\mathrm{V}$, the new power factor of the circuit will be:

Related Formula

For a series RC branch, the voltage phase angle θ$\theta$ is:

If the phase difference between Vᵢₙ$V_{\text{in}}$ and (VB - VA)$(V_B - V_A)$ is 90^°$90^\circ$:

Core Logic

• The phase angle of VA$V_A$ is θ$\theta$ behind the input voltage.
• The phase angle of VB$V_B$ is 90^° - θ$90^\circ - \theta$ ahead of the input.
• If their combined relative phase difference is 90^°$90^\circ$, this symmetry dictates:

Therefore, we have the condition:

Step 1: Solve for Frequency

Rearrange to solve for ω$\omega$:

Substitute the given values:

• R = 100 ~kΩ = 10⁵ Ω$R = 100 \mathrm{~k}\Omega = 10^5 \Omega$
• C = 100 ~pF = 100 × 10⁻¹² ~F = 10⁻¹⁰ ~F$C = 100 \mathrm{~pF} = 100 \times 10^{-12} \mathrm{~F} = 10^{-10} \mathrm{~F}$

Since ω = 10^x$\omega = 10^x$, we get x = 5$x = 5$.

Pattern Recognition

Sees: Phase shift of 90^°$90^\circ$ across a symmetric RC bridge. Shortcut: Clamping phase shift at 90^°$90^\circ$ implies the reactive impedance equals the resistive impedance (XC = R$X_C = R$). The frequency is simply the characteristic time constant frequency ω = 1/τ = 1/(RC)$\omega = 1/\tau = 1/(RC)$.

Chapter Mix

Class 12 Physics: Alternating Current

Related Formula

For a current having both DC and AC components i = Idc + I₀ (ω t + φ)$i = I_{\text{dc}} + I_0 \cos(\omega t + \phi)$, the mean-square value is:

The RMS current Irms$I_{\text{rms}}$ is:

Core Logic

Identify the parameters from the given equation:

• Idc = 5√(2) ~A$I_{\text{dc}} = 5\sqrt{2} \mathrm{~A}$
• I₀ = 10 ~A$I_0 = 10 \mathrm{~A}$

Calculate the square of the components:

Step 1: Compute Resultant RMS

Substitute back into the RMS formula:

Pattern Recognition

Sees: Superposition of a DC current Idc$I_{\text{dc}}$ and a pure AC cosine wave with amplitude Iac$I_{\text{ac}}$. Shortcut: Use the orthogonal component RMS formula: Irms = Idc² + Iac,rms²$I_{\text{rms}} = \sqrt{I_{\text{dc}}^2 + I_{\text{ac,rms}}^2}$. Here Idc = 5√(2)$I_{\text{dc}} = 5\sqrt{2}$ and Iac,rms = 10/√(2) = 5√(2)$I_{\text{ac,rms}} = 10/\sqrt{2} = 5\sqrt{2}$. Thus, Irms = √(50 + 50) = 10 ~A$I_{\text{rms}} = \sqrt{50 + 50} = 10 \mathrm{~A}$.

Chapter Mix

Class 12 Physics: Alternating Current

Related Formula

Inductive Reactance:

Impedance of RL series circuit:

Maximum current:

Core Logic

Calculate inductive reactance XL$X_L$:

Given resistance R = 100π Ω$R = 100\pi\ \Omega$. Compute total impedance Z$Z$:

Step 1: Calculate Maximum Current

Assuming the given supply voltage (100π V$100\pi\text{ V}$) is standard RMS voltage:

Then, peak/maximum current is:

Hence, the maximum current is 1.

Pattern Recognition

When inductive reactance equals resistance (XL = R$X_L = R$), the impedance is exactly R√(2)$R\sqrt{2}$. The factor of √(2)$\sqrt{2}$ in the denominator cancels perfectly with the peak current conversion multiplier.

Chapter Mix

Class 12 Physics: Alternating Current

Related Formula

Therefore:

where:

• β$\beta$ = fringe width
• d$d$ = slit separation width
• D$D$ = distance to screen
• λ$\lambda$ = wavelength

Core Logic

Given data:

• Initial slit separation, d₁ = 0.2~mm$d_1 = 0.2\mathrm{~mm}$
• Final slit separation, d₂ = 0.4~mm$d_2 = 0.4\mathrm{~mm}$ (d$d$ is doubled).

Step 1: Calculate Percentage Change

Since d₂ = 2d₁$d_2 = 2d_1$, the new fringe width becomes:

Percentage change formulation:

Pattern Recognition

Doubling the denominator of an inversely proportional relationship halves the primary value, yielding an absolute 50%$50\%$ decrease.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Z = √(R² + (XL - XC)²)$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ [cite: 786]

P = Irms² R = Vrms² RZ²$$P = I_{\text{rms}}^2 R = \frac{V_{\text{rms}}^2 R}{Z^2}$$ [cite: 785, 786]

Core Logic

First, find the total impedance Z$Z$ of the LCR circuit: [cite: 185, 786]

Z = √(50² + (100 - 50)²) = √(50² + 50²) = 50√(2) Ω$$Z = \sqrt{50^2 + (100 - 50)^2} = \sqrt{50^2 + 50^2} = 50\sqrt{2}\ \Omega$$ [cite: 185, 786]

Now calculate the average power dissipation: [cite: 185, 786]

P = (10)² × 50(50√(2))² = (100 × 50)/(2500 × 2) = (5000)/(5000) = 1 W$$P = \frac{(10)^2 \times 50}{(50\sqrt{2})^2} = \frac{100 \times 50}{2500 \times 2} = \frac{5000}{5000} = 1\ \text{W}$$ [cite: 787]

Pattern Recognition

Remember that only the resistive element dissipates real thermal power over a complete cycle[cite: 784, 785]. Reactive elements like ideal inductors and capacitors store and release energy alternately without net consumption.

Chapter Mix

Class 12 Physics: Alternating Current