A capacitor of capacitance 1 μ F is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance 10- 3 H. The maximum current that will flow in the circuit has the value

Question

A capacitor of capacitance 1 μ F is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance 10- 3 H. The maximum current that will flow in the circuit has the value (A)  √1000 mA (B) 1 A (C) 1 mA (D) 1000 mA

Tardigrade Admin · Accepted Answer

Change on the capacitor, q0 = CV = 1 × 10 - 6 × 1 = 10 - 6 C here, q = q0 sin ω t or I0 = ω q0 = maximum current Now, ω = ( 1/ √ LC) = ( 1/ √ 10 - 9) = ( 109 ) 1/2 ∴ i0 = ( 109)1/2 × (1 × (10 - 6 ) = √1000 mA

Q. A capacitor of capacitance $1 μ F$ is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance $1 0^{- 3}$ H. The maximum current that will flow in the circuit has the value

$1000$ mA

1 A

1 mA

1000 mA

Change on the capacitor,
$q_{0} = C V = 1 \times 1 0^{- 6} \times 1 = 1 0^{- 6}$ C
here, $q = q_{0} sin ω t$
or $I_{0} = ω q_{0} =$ maximum current
Now, $ω = \frac{1}{L C} = \frac{1}{1 0 ^{- 9}} = (1 0^{9})^{1/2}$
$∴ i_{0} = (1 0^{9})^{1/2} \times (1 \times (1 0^{- 6})$
$= 1000$ mA

Q. A capacitor of capacitance 1μF is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance 10−3 H. The maximum current that will flow in the circuit has the value

1000​ mA

1 A

1 mA

1000 mA

Change on the capacitor, q0​=CV=1×10−6×1=10−6 C here, q=q0​sinωt or I0​=ωq0​= maximum current Now, ω=LC​1​=10−9​1​=(109)1/2 ∴i0​=(109)1/2×(1×(10−6) =1000​ mA

Q. A capacitor of capacitance $1 μ F$ is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance $1 0^{- 3}$ H. The maximum current that will flow in the circuit has the value

$1000$ mA

Change on the capacitor,
$q_{0} = C V = 1 \times 1 0^{- 6} \times 1 = 1 0^{- 6}$ C
here, $q = q_{0} sin ω t$
or $I_{0} = ω q_{0} =$ maximum current
Now, $ω = \frac{1}{L C} = \frac{1}{1 0 ^{- 9}} = (1 0^{9})^{1/2}$
$∴ i_{0} = (1 0^{9})^{1/2} \times (1 \times (1 0^{- 6})$
$= 1000$ mA