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7.Alternating Current
hard
Let $f = 50\, Hz$, and $C = 100\, \mu\, F$ in an $AC$ circuit containing a capicator only. If the peak value of the current in the circuit is $1.57$ $A$ at $t = 0$. The expression for the instantaneous voltage across the capacitor will be
A
$E = 50\, sin (100\, \pi t - \pi /2)$
B
$E = 100\, sin (50\, \pi t)$
C
$E = 50 \,sin\, 100\, \pi t$
D
$E = 50\, sin\, (100\, \pi\, t + \pi /2)$
Solution
$f=50 H z, C=100 \mu F X_{C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times \pi \times 50 \times 100 \times 10^{-6}}=31.85 \Omega$
$I_{m}=1.57 A$ at $t=0$
Instantaneous voltage is $E=E_{o} \sin w t$
$E_{o}=X_{C} \times I=31.85 \times 1.57=V$
$\therefore E=\sin \left(w e-90^{\circ}\right)$
$\therefore E=50 \sin (100 \pi t+\pi / 2)$
Standard 12
Physics
