Q29. A e.m.f. es = 50sin 314t is applied across a pure capacitor of 637µF. The instantaneous current

Q: What is the correct answer for CUET UG Physics Q29?

The correct answer is a. For a pure capacitive AC circuit, the instantaneous current (I) leads the instantaneous EMF (E) by a phase angle of π/2 (or 90°). If the EMF is given by E = E₀sin(ωt), then the current will be I = I₀sin(ωt + π/2) = I₀cos(ωt). First, identify E₀ and ω from the given EMF: E = 50sin(314t), so E₀ = 50 V and ω = 314 rad/s. The capacitance C = 637 µF = 637 × 10⁻⁶ F. Next, calculate the capacitive reacta

Q: Q29. A e.m.f. es = 50sin 314t is applied across a pure capacitor of 637µF. The instantaneous current I is: (a) 10cos 314tA (b) 50cos 314tA (c) 20cos 314tA (d) 20sin 314tA

The correct answer is a. For a pure capacitive AC circuit, the instantaneous current (I) leads the instantaneous EMF (E) by a phase angle of π/2 (or 90°). If the EMF is given by E = E₀sin(ωt), then the current will be I = I₀sin(ωt + π/2) = I₀cos(ωt). First, identify E₀ and ω from the given EMF: E = 50sin(314t), so E₀ = 50 V and ω = 314 rad/s. The capacitance C = 637 µF = 637 × 10⁻⁶ F. Next, calculate the capacitive reacta

Q: What are the options for this CUET UG Physics question?

A. 10cos 314tA · B. 50cos 314tA · C. 20cos 314tA · D. 20sin 314tA

Question

Q29. A e.m.f. es = 50sin 314t is applied across a pure capacitor of 637µF. The instantaneous current I is:
(a) 10cos 314tA
(b) 50cos 314tA
(c) 20cos 314tA
(d) 20sin 314tA

FirstInTest · Accepted Answer

For a pure capacitive AC circuit, the instantaneous current (I) leads the instantaneous EMF (E) by a phase angle of π/2 (or 90°). If the EMF is given by E = E₀sin(ωt), then the current will be I = I₀sin(ωt + π/2) = I₀cos(ωt). First, identify E₀ and ω from the given EMF: E = 50sin(314t), so E₀ = 50 V and ω = 314 rad/s. The capacitance C = 637 µF = 637 × 10⁻⁶ F. Next, calculate the capacitive reactance (X_C): X_C = 1/(ωC) = 1/(314 × 637 × 10⁻⁶). Since 314 is approximately 100π (3.14 x 100), and 637 x 10⁻⁶ is approximately 2/π x 10⁻³ (2/3.14 x 10⁻³), XC is approximately 1/(100π * 637 * 10⁻⁶) which is close to 5 Ω. Let's use 314 * 637 * 10⁻⁶ ≈ 0.2, so X_C ≈ 1/0.2 = 5 Ω. Now, calculate the peak current I₀ = E₀/X_C = 50 V / 5 Ω = 10 A. Therefore, the instantaneous current is I = 10cos(314t) A, matching option (a).

Answer

10cos 314tA

Answer

50cos 314tA

Answer

20cos 314tA

Answer

20sin 314tA

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