Physics Alternating Current

Diameter = main scale reading + (circular scale reading * least count)
Diameter = 0 + (52 * 0.01 mm) = 0.52 mm = 0.052 cm.
In the RLC circuit:
Given I? = 10√2 A, so I? = I? /√2 = 10 A.
V? = √ [V? ² + (V? - V? )²] = √ [40² + (40 - 10)²] = √ [1600 + 30²] = √ [1600 + 900] = √2500 = 50 V.
Impedance Z = V? / I? = 50 V / 10 A = 5 Ω.
For Hindi: I? = 10√2 A, V? = 50V, Z = V? /I? = 50/ (10√2) = 5/√2 Ω.

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{48}{3*{10}^8}=16*{10}^{-8}$

$=1.6*{10}^{-7}\mathrm{T}$

Peak voltage is $\sqrt[]{2}$  times rms voltages in ac.

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{48}{3*{10}^8}=16*{10}^{-8}$

$=1.6*{10}^{-7}\mathrm{T}$

Capacitive reactance $=\frac{1}{\omega \mathrm{C}}={\mathrm{X}}_{\mathrm{c}}$ (say) on decreasing the operating frequency $\omega$  reduces

As $X_c$ is inversely proportional to $\omega$ the value of $X_c$  increase

$\therefore {\mathrm{I}}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{D}}$

$=\frac{{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{c}}}$

As ${\mathrm{X}}_{\mathrm{c}}$  increases, therefore displacement current Id decreases.

z = √ [R² + (X? - X? )²] = √ [6² + (4-10)²] = 6√2 Ω

Power factor = cosφ = R/z = 6/ (6√2) = 1/√2

T = 2π√ (l/g) ⇒ g = 4π²l/T²
Percentage error: Δg/g = Δl/l + 2 (ΔT/T) = (0.1/10.0) + 2 (0.005/0.5) = 0.03
Percentage error = (Δg/g) * 100 = 3%
ω = 2πf = 100π rad/s
i_rms = i? /√2
While current changes from its maximum to its rms value, its phase changes by π/4 rad.
t = (π/4)/ω = π/ (4 * 100π) = 2.5 * 10? ³ s = 2.5ms.

Resonance frequency is independent of R.
Quality factor = ωL/R ⇒ Quality factor decreases with increase in R.
Bandwidth of resonance circuit = R/L ⇒ increases with increase in R.

I = I? sin (ωt) + I? cos (ωt)

This can be written as I = I? sin (ωt + φ), where I? = √ (I? ² + I? ²)

A hot wire ammeter reads the rms value of the current.

I_rms = I? /√2 = √ (I? ² + I? ²)/√2