Question

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Answer 1

Initial the current flows through 2 ×10^-6 F capacitor only

Initially the charge stored in the capacitor is given by the formula

$q = cv$

$q = 2v \times {10}^{ - 6}$

Initially the energy stored in the capacitor is given by the formula

$u = \frac{1c {v}^{2} }{2}$

$u = \frac{1 \times 2 \times {10}^{ - 6} \times {v}^{2} }{2}$

$u = {v}^{2} \times {10}^{ - 6}$

when the switch is turned to position 2 current flows through both the capacitors

As both the capacitors are connected in parallel resultant capacitance is

C = 8+2 ×10^-6

C = 10^-5

Q =2 v ×10^-6

V = Q/C

V = 2 v ×10^-6/10^-5

V = 2 v /10

V =V /5

U final = 1/2 10^-5 v²/25

U final = v² ×10^-5/50

U final = v²/5 ×10 ^-6

Energy dissipated

= U - U f /U ×100

= V²-V²/5 /V ×100

= 4 V²/5/V ×100

= 4 V /5 ×100

= 80 %