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Let frequency \mathrm{v=50 \mathrm{~Hz}}, and capacitance \mathrm{\mathrm{C}=100 \mu \mathrm{F}} in an ac circuit containing a capacitor only. If the peak value of the current in the circuit is 1.57 A. The expression for the instantaneous voltage across the capacitor will be

Option: 1

\mathrm{E=50 \sin \left(100 \pi t-\frac{\pi}{2}\right)}


Option: 2

\mathrm{E=100 \sin (50 \pi t)}


Option: 3

\mathrm{E=50 \sin (100 \pi t)}


Option: 4

\mathrm{E=50 \sin \left(100 \pi t+\frac{\pi}{2}\right)}


Answers (1)

best_answer

\mathrm{\text { Peak value of voltage } V_0=i_0 X_C=\frac{i_0}{2 \pi v C} \Rightarrow \frac{1.57}{2 \times 3.14 \times 50 \times 100 \times 10^{-6}}=50 \mathrm{~V}}

\mathrm{\text { Hence if equation of current } \mathrm{i}=\mathrm{i}_0 \sin \omega \mathrm{t} \text { then in capacitive circuit voltage is }}

\mathrm{\begin{aligned} & V=V_0 \sin \left(\omega t-\frac{\pi}{2}\right) \\ & \Rightarrow V=50\left(\sin 2 \pi \times 50 t-\frac{\pi}{2}\right)=50 \sin \left(100 \pi t-\frac{\pi}{2}\right) \end{aligned}}

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