An alternating potential V=V₀ sin wt is applied across a circuit. As a result the current I=I₀ Sin $I=I_0 sin \left(\omega t-\frac{\pi }{2}\right)$  flows in it. The power consumed in the circuit per cycle is

$\omega =2\pi \mathrm{f}\Rightarrow \omega =100\pi$

$\mathrm{Z}=\sqrt[]{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$

$=\sqrt[]{{10}^2+{\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)}^2}$

$=\sqrt[]{100+{\left(100\pi \times \frac{50}{\pi }\times {10}^{-3}-\frac{1}{100\pi \times \frac{{10}^3}{\pi }\times {10}^{-6}}\right)}^2}$

$=\sqrt[]{100+(5-10{)}^2}$

$=\sqrt[]{100+25}$

$\mathrm{Z}=5\sqrt[]{5}\mathrm{\Omega }$

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

X_C  = 1/ ωC

On increasing w; XC decreases so Z decreases, current in circuit increases.

$As{V}^2=V_R^2+C_C^2$

$?V_R=\sqrt[]{V^2-V_C^2}=\sqrt[]{{10}^2-8^2}=6V$

$\tan ?\theta =\frac{X_C}{X_R}=\frac{V_C}{V_R}=\frac{8}{6}=\frac{4}{3}$  

Now,