Question

Part A An L-C circuit containing an 88.0-mH inductor and a 1.45-nF capacitor oscillates with a maximum current of 0.710 A. You may want to review (Page). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of An oscillating circuit. Calculate the maximum charge on the capacitor. Q = 8.02x10<sup>-6</sup> C Submit Previous Answers Part B Correct IDENTIFY: Apply energy conservation. SET UP: If I is the maximum current, $\frac{1}{2}LI^2 = \frac{Q^2}{2C}$. EXECUTE: $\frac{1}{2}LI^2 = \frac{Q^2}{2C}$ gives $Q = I\sqrt{LC} = (0.710\ A)\sqrt{(88.0\ mH)(1.45\ nF)} = 8.02 \times 10^{-6}\ C$. Calculate the oscillation frequency of the circuit. f = 1.41x10<sup>4</sup> Hz Submit Previous Answers ? Correct Correct answer is shown. Your answer 14089 Hz was either rounded differently or used a different number of significant figures than required for this part. $\omega = \frac{1}{\sqrt{LC}} = \sqrt{(88.0\ mH)(1.45\ nF)} = 8.853 \times 10^4\ rad/s \approx 8.85 \times 10^4\ rad/s. f = \frac{\omega}{2\pi} = 1.41 \times 10^4\ Hz$. Part C Assuming the capacitor had its maximum charge at time t = 0, calculate the energy stored in the inductor after 2.55 ms of oscillation.

          Part A
An L-C circuit containing an 88.0-mH inductor and a
1.45-nF capacitor oscillates with a maximum current of 0.710
A.
You may want to review (Page).
For related problemsolving tips and strategies, you may want to
view a Video Tutor Solution of An oscillating circuit.
Calculate the maximum charge on the capacitor.
Q = 8.02x10<sup>-6</sup> C
Submit Previous Answers
Part B
Correct
IDENTIFY: Apply energy conservation.
SET UP: If I is the maximum current, $\frac{1}{2}LI^2 = \frac{Q^2}{2C}$.
EXECUTE: $\frac{1}{2}LI^2 = \frac{Q^2}{2C}$ gives $Q = I\sqrt{LC} = (0.710\ A)\sqrt{(88.0\ mH)(1.45\ nF)} = 8.02 \times 10^{-6}\ C$.
Calculate the oscillation frequency of the circuit.
f = 1.41x10<sup>4</sup> Hz
Submit Previous Answers
? Correct
Correct answer is shown. Your answer 14089 Hz was either rounded differently or used a different number of significant figures than required
for this part.
$\omega = \frac{1}{\sqrt{LC}} = \sqrt{(88.0\ mH)(1.45\ nF)} = 8.853 \times 10^4\ rad/s \approx 8.85 \times 10^4\ rad/s. f = \frac{\omega}{2\pi} = 1.41 \times 10^4\ Hz$.
Part C
Assuming the capacitor had its maximum charge at time t = 0, calculate the energy stored in the inductor after 2.55 ms of oscillation.

Part A
An L-C circuit containing an 88.0-mH inductor and a
1.45-nF capacitor oscillates with a maximum current of 0.710
A.
You may want to review (Page).
For related problemsolving tips and strategies, you may want to
view a Video Tutor Solution of An oscillating circuit.
Calculate the maximum charge on the capacitor.
Q = 8.02x10<sup>-6</sup> C
Submit Previous Answers
Part B
Correct
IDENTIFY: Apply energy conservation.
SET UP: If I is the maximum current, (1)/(2)LI^2 = (Q^2)/(2C).
EXECUTE: (1)/(2)LI^2 = (Q^2)/(2C) gives Q = I√(LC) = (0.710 A)√((88.0 mH)(1.45 nF)) = 8.02 × 10^-6 C.
Calculate the oscillation frequency of the circuit.
f = 1.41x10<sup>4</sup> Hz
Submit Previous Answers
? Correct
Correct answer is shown. Your answer 14089 Hz was either rounded differently or used a different number of significant figures than required
for this part.
ω = (1)/(√(LC)) = √((88.0 mH)(1.45 nF)) = 8.853 × 10^4 rad/s ≈ 8.85 × 10^4 rad/s. f = (ω)/(2π) = 1.41 × 10^4 Hz.
Part C
Assuming the capacitor had its maximum charge at time t = 0, calculate the energy stored in the inductor after 2.55 ms of oscillation.

Added by Michael H.

Question

Please give Ace some feedback