x= ______________ % In the figure, the battery has an emf of 11.0 V, the inducta

Question

x= ______________ %

In the figure, the battery has an emf of 11.0 V, the inductance is 2.96 mH, and the capacitance is 9.00 pF. The switch has been set to position a for a long time so that the capacitor is charged. The switch is then thrown to position b, removing the battery from the circuit and connecting the capacitor directly across the inductor. First the capacitor is fully charged with the switch set to position a. Then, the switch is thrown to position b and the battery is no longer in the circuit. What if the charge on the capacitor is found to be 84 % of the maximum value at a particular instant, what is the current (magnitude only) flowing through the circuit at that instant? Give your answer as a percentage to the maximum current value.

Explanation / Answer

total charge stored in capacitor is q=cv which is equal to q(max)=99pc

hence this is the maximum charge. so 84% of maximum charge is q = 84x99/100 pc = 83.16pc

hence when it is connected across an inductor then energy is conserved

so q2/(2c) + li2 /2 = q(max)2 /(2c)

hence substituting the values we get i=0.329 milliamp

maximum current is calculated by q(max)2 /(2c) = li(max)2 /2

which is i(max) = 0.604 milli amp

so in terms of percentage its 54.46%

Related Questions

Navigate

• Browse (All)
• Browse X
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.