Question

An oscillating LC circuit consisting of a 1.0 nF capacitor and a 2.4 mH coil has a maximum voltage of 2.5 V. What are (a) the maximum charge on the capacitor, (b) the maximum current through the circuit, (c) the maximum energy stored in the magnetic field of the coil

Answer 1

a.
`q_m=2.5* 10^(-6)\ C`

b.
`i_m=5.1* 10^(-2)\ A`

c.
`U_B=3.1212* 10^(-6)\ J`

Step-by-step explanation:

Given:

• capacitance,
  `C=10^(-6)\ F`

• maximum voltage,
  `V_m=2.5\ V`

• inductance,
  `L=2.4* 10^(-3) \ H`

(a)

maximum charge on the capacitor:

`q_m=C.V_m`

`q_m=10^(-6)* 2.5`

`q_m=2.5* 10^(-6)\ C`

(b)

∵Energy store in electric and magnetic fields are equal:

`\therefore U_E=U_B`

`(1)/(2) * (q_m)/(C) =(1)/(2)* L.i_m^2`

`i_m=(q_m)/(√(L.C) )`

`i_m=\frac{2.5* 10^(-6)}{\sqrt{2.4* 10^(-3)* 10^(-6)} }`

`i_m=5.1* 10^(-2)\ A`

(c)

The maximum energy stored in the magnetic field of the coil:

`U_B=(1)/(2)* L.i_m^2`

`U_B=(1)/(2)* 2.4* 10^(-3)* (5.1* 10^(-2))^2`

`U_B=3.1212* 10^(-6)\ J`

Answer 2

`2.5* 10^(-9)\ C`

0.00161 A

`3.125* 10^(-9)\ J`

Step-by-step explanation:

C = Capacitance = 1 nF

L = Inductance = 2.4 mH

`V_m` = V = Voltage = 2.5 V

I = Current

Maximum charge in capacitor is given by

`Q_m=CV_m\\\Rightarrow Q_m=1* 10^(-9)* 2.5\\\Rightarrow Q_m=2.5* 10^(-9)\ C`

The maximum charge on the capacitor is
`2.5* 10^(-9)\ C`

Energy stored in capacitor is given by

`U=(1)/(2)CV^2\\\Rightarrow U=(1)/(2)* 1* 10^(-9)* 2.5^2\\\Rightarrow U=3.125* 10^(-9)\ J`

The maximum energy stored in the magnetic field of the coil is
`3.125* 10^(-9)\ Joules`

Energy stored in coil is given by

`U=(1)/(2)LI^2\\\Rightarrow I=\sqrt{(2U)/(L)}\\\Rightarrow I=\sqrt{(2* 3.125* 10^(-9))/(2.4* 10^(-3))}\\\Rightarrow I=0.00161\ A`

The current passing through the circuit is 0.00161 A