A 1.0 μ F capacitor is charged by a 40.0 V power supply. The fully charged capacitor is then discharged through a 10.0- mH inductor. The maximum current in the resulting oscillations (in mA ) is

Question

A 1.0 μ F capacitor is charged by a 40.0 V power supply. The fully charged capacitor is then discharged through a 10.0- mH inductor. The maximum current in the resulting oscillations (in mA ) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

At different times, (UC) max =(UL) max , so [(1/2) C(Δ V)2] max =(1/2) L li2 Then, Ii=√(C/L)(Δ V) max =√(1.00 × 10-6 F /10.0 × 10-3 H )(40.0 V ) =0.400 A =400 mA

Q. A $1.0 μ F$ capacitor is charged by a $40.0 V$ power supply. The fully charged capacitor is then discharged through a $10.0 - m H$ inductor. The maximum current in the resulting oscillations (in $m A$ ) is___

At different times, $(U_{C})_{m a x} = (U_{L})_{m a x}$ ,
so $[\frac{1}{2} C (Δ V)^{2}]_{m a x} = \frac{1}{2} L l_{i}^{2}$
Then, $I_{i} = \frac{C}{L} (Δ V)_{m a x}$
$= \frac{1.00 \times 1 0 ^{- 6} F}{10.0 \times 1 0 ^{- 3} H} (40.0 V)$
$= 0.400 A = 400 m A$

Q. A 1.0μF capacitor is charged by a 40.0V power supply. The fully charged capacitor is then discharged through a 10.0−mH inductor. The maximum current in the resulting oscillations (in mA ) is___

At different times, (UC​)max​=(UL​)max​, so [21​C(ΔV)2]max​=21​Lli2​ Then, Ii​=LC​​(ΔV)max​ =10.0×10−3H1.00×10−6F​​(40.0V) =0.400A=400mA

Q. A $1.0 μ F$ capacitor is charged by a $40.0 V$ power supply. The fully charged capacitor is then discharged through a $10.0 - m H$ inductor. The maximum current in the resulting oscillations (in $m A$ ) is___

At different times, $(U_{C})_{m a x} = (U_{L})_{m a x}$ ,
so $[\frac{1}{2} C (Δ V)^{2}]_{m a x} = \frac{1}{2} L l_{i}^{2}$
Then, $I_{i} = \frac{C}{L} (Δ V)_{m a x}$
$= \frac{1.00 \times 1 0 ^{- 6} F}{10.0 \times 1 0 ^{- 3} H} (40.0 V)$
$= 0.400 A = 400 m A$