A series circuit consists of a resistor $R = 16 \Omega$, an inductor with $L = 2 H$, a capacitor with $C = 0.006250 F$, and a $12 - V$ battery. If the initial charge is $0.0008 \mathrm { C }$ and the initial current is 0 , find the current $I ( t )$ at time $t$.
A)$I ( t ) = \frac { e ^ { - 4 t } } { 160 } ( 0.006250 \cos ( 4 t ) + 8 \sin ( 4 t ) ) - \frac { 1 } { 2 }$
B) $I ( t ) = 0.742 e ^ { - 4 t } \sin ( 8 t )$
C)$I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t ) \frac { 1 } { 2 }$
D)$I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t )$
E) $I ( t ) = 0.742 e ^ { - 8 t }$

Question

A series circuit consists of a resistor $R = 16 \Omega$, an inductor with $L = 2 H$, a capacitor with $C = 0.006250 F$, and a $12 - V$ battery. If the initial charge is $0.0008 \mathrm { C }$ and the initial current is 0 , find the current $I ( t )$ at time $t$.
A)$I ( t ) = \frac { e ^ { - 4 t } } { 160 } ( 0.006250 \cos ( 4 t ) + 8 \sin ( 4 t ) ) - \frac { 1 } { 2 }$
B) $I ( t ) = 0.742 e ^ { - 4 t } \sin ( 8 t )$
C)$I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t )  \frac { 1 } { 2 }$
D)$I ( t ) = \frac { e ^ { - 4 t } } { 160 } \cos ( 8 t )$
E) $I ( t ) = 0.742 e ^ { - 8 t }$

Quizplus · Accepted Answer

The given circuit is an RLC series circuit. The solution for the current in such a circuit with given initial conditions is typically of the form $I(t) = A e^{-\alpha t} \sin(\omega_d t)$, where $\alpha = \frac{R}{2L}$ and $\omega_d = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}$. For the given values, $\alpha = 4$ and $\omega_d = 8$, which matches option B.

A Series Circuit Consists of a Resistor R=16ΩR = 16 \OmegaR=16Ω

A Series Circuit Consists of a Resistor $R = 16 \Omega$