Question

In the circuit of the figure below, the battery emf E is 50 V, the resistance R is 210 Q, and the capacitance C is 0.500 pF. The switch S is closed for a long time interval, and zero potential difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum value of 150 V. What is the value of the inductance? Н С

Answer 1

R= 2lor VA &=500 = 0.5 uf R 250 Initially, when swhich s is closed for long time. current (1) will be I = &-50A energy stored in inductor will be E= 1 L 1 ² When switch is open as in fig (B), the energy stored in inductor will charge capacitor up to potential dift. v= 150v so energy stored in inductor will convent into energy stored in capacitor, when capacities is charged up to v=1500, by energy conservation LI? Len? I- & - SOA ( 58 ) ² = 0.5 x 15 x 11503² = 0.5 UF L = 0.19845H C = 0.5x10 of L = 0.20 H