Initially the capacitor is charged to a potential of 5V and then connected to position 1 with the shown polarity for 1s. After 1s, it is connected across the inductor at position 2.
Choose correct alternative(s)
A
The potential across capacitor after 1s of its connection to position (1) is 5[2−1e]V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The maximum current flowing in the L−C circuit when the capacitor is connected across the inductor is 10[2−1e]A
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The frequency of LC oscillation is 100π Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C The frequency of LC oscillation is 100π Hz Position (1)-
For RC circuit,
Using KVL, E−IR−qC=0 E−dqdtR−qC=0 dqdtR=E−qC RdqE−qC=dt
Integrationg above equation within limits ∫qq0RdqE−qC=∫10dt
R⎡⎢
⎢
⎢
⎢⎣ln(E−qC)−1C⎤⎥
⎥
⎥
⎥⎦qq0=[t]10
ln(E−qC)−ln(E−q0C)=−tRC
Here q0 is initial charge, given that initially capacitor is charged to 5V, hence q0C=5V
Also, E=10V,C=10 mF,R=100Ω,t=1s
Substituting the values we get, ln(10−qC)−ln(10−5)=−1100(10×10−3) ln(2−q5C)=−1 2−q5C=1e qC=5[2−1e]
Hence potential across capacitor at t=1 is V=qC=5[2−1e]V
Position(2)−
→During L−C oscillations, when current in it is maximum. Total energy will be stored in the inductor (12LI2),
Using conservation of energy, 12CV2=12LI2 12(10×10−3)[5(2−1e)]2=12×2.5×10−3×I2 ⇒Imax=10(2−1e)A
Frequency of LC oscillation f=12π√LC=12π1√2.5×10×10−6⇒f=100π Hz a,b,c are correct