Question

a capacitor of capacitance 1 microfarad is charged to a potential of 1 volt it is connected in parallel to an inductor of inductance 1 Milli Henry the maximum current That Flow in the circuit has the value​

Answer 1

Answer:

1/2CV^2=1/2Li^2

i^2=CV^2/L

     =10^-6*1/10^-3

       =10^-3

Answer 2

The maximum current that flow in the circuit has the value​ 1 mA.

Explanation:

As circuit is LC, and no resistance in circuit so the maximum energy stored in the capacitor is equal to the maximum energy stored in the inductor and given as

$U_C=U_L$

$\frac{1}{2}CV^2=\frac{1}{2}Li^2$

Here, C is the capacitance, V is the voltage, L is the inductance and i is the maximum current in LC circuit.

Given C = 1 microfarad, V = 1 V, L =  1 mH.

Substitute the given values, we get

 $\frac{1}{2}\times1\times10^-^6F=\frac{1}{2}\times1\times10^-^3\times i$

$i=\frac{10^-^6}{10^-^3}=10^-^3A$

$i=1mA$

Thus, the maximum current that flow in the circuit has the value​ 1 mA.

#Learn More: LC circuit.

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