A capacitor of capacitance 150 . 0   μ F is connected to an alternating source of emf given by E = 36   sin 120 π t   V . The maximum value of current in the circuit is approximately equal to:

Question

A capacitor of capacitance 150 . 0 &#160; &#956; F is connected to an alternating source of emf given by E = 36 &#160; sin 120 &#960; t &#160; V . The maximum value of current in the circuit is approximately equal to:

Aicharya · Accepted Answer

The formula for maximum current is given by
$I_{m a x} = ω C V_{m} . . . (i)$
The given data is
$ω = 120 π$
$V_{m} = 36 V C = 150 μF$
Substituting the values in equation (i)
$I_{\max} = (120 π) \times (150 \times 10^{- 6}) \times 36$
$= 2.036 A$
$\approx 2 A$

A capacitor of capacitance 150 . 0   μ F is connected to an alternating source of emf given by E =…

Solution

A capacitor of capacitance 150 . 0 &#160; &#956; F is connected to an alternating source of emf given by E =…

Solution

A capacitor of capacitance 150 . 0 μ F is connected to an alternating source of emf given by E =…