Respuesta :
Answer:
[tex]2.5\times 10^{-9}\ C[/tex]
0.00161 A
[tex]3.125\times 10^{-9}\ J[/tex]
Explanation:
C = Capacitance = 1 nF
L = Inductance = 2.4 mH
[tex]V_m[/tex] = V = Voltage = 2.5 V
I = Current
Maximum charge in capacitor is given by
[tex]Q_m=CV_m\\\Rightarrow Q_m=1\times 10^{-9}\times 2.5\\\Rightarrow Q_m=2.5\times 10^{-9}\ C[/tex]
The maximum charge on the capacitor is [tex]2.5\times 10^{-9}\ C[/tex]
Energy stored in capacitor is given by
[tex]U=\frac{1}{2}CV^2\\\Rightarrow U=\frac{1}{2}\times 1\times 10^{-9}\times 2.5^2\\\Rightarrow U=3.125\times 10^{-9}\ J[/tex]
The maximum energy stored in the magnetic field of the coil is [tex]3.125\times 10^{-9}\ Joules[/tex]
Energy stored in coil is given by
[tex]U=\frac{1}{2}LI^2\\\Rightarrow I=\sqrt{\frac{2U}{L}}\\\Rightarrow I=\sqrt{\frac{2\times 3.125\times 10^{-9}}{2.4\times 10^{-3}}}\\\Rightarrow I=0.00161\ A[/tex]
The current passing through the circuit is 0.00161 A
Answer:
a. [tex]q_m=2.5\times 10^{-6}\ C[/tex]
b. [tex]i_m=5.1\times 10^{-2}\ A[/tex]
c. [tex]U_B=3.1212\times 10^{-6}\ J[/tex]
Explanation:
Given:
- capacitance, [tex]C=10^{-6}\ F[/tex]
- maximum voltage, [tex]V_m=2.5\ V[/tex]
- inductance, [tex]L=2.4\times 10^{-3} \ H[/tex]
(a)
maximum charge on the capacitor:
[tex]q_m=C.V_m[/tex]
[tex]q_m=10^{-6}\times 2.5[/tex]
[tex]q_m=2.5\times 10^{-6}\ C[/tex]
(b)
∵Energy store in electric and magnetic fields are equal:
[tex]\therefore U_E=U_B[/tex]
[tex]\frac{1}{2} \times \frac{q_m}{C} =\frac{1}{2}\times L.i_m^2[/tex]
[tex]i_m=\frac{q_m}{\sqrt{L.C} }[/tex]
[tex]i_m=\frac{2.5\times 10^{-6}}{\sqrt{2.4\times 10^{-3}\times 10^{-6}} }[/tex]
[tex]i_m=5.1\times 10^{-2}\ A[/tex]
(c)
The maximum energy stored in the magnetic field of the coil:
[tex]U_B=\frac{1}{2}\times L.i_m^2 [/tex]
[tex]U_B=\frac{1}{2}\times 2.4\times 10^{-3}\times (5.1\times 10^{-2})^2[/tex]
[tex]U_B=3.1212\times 10^{-6}\ J[/tex]
