capacitorcurrent-sourcevoltage-source
I came upon this sentence in a paper concerned with dielectric elastomer actuators (DEA):
Due to the capacitive behavior of a DE, a converter with current source functionality should, therefore, preferably be used to adjust the voltage vp, and by this the electrostatic pressure, in (1).
Why is a current source preferable compared to a voltage source?
By what i can see you diagram is wrong. To convert the two current sources you convert the sources considering the parallel resistors as the figure below.
To convert the current source to a voltage source you first substitute the pair (current source)-(resistor in parallel) by a series association of a voltage source and a resistor with the same value as the resistor in parallel. The value of the voltage source is given by the formula V=R*I. Pay attention to the direction ofthe voltage source as it needs to have the same direction as the current source.
Applying the transformation leads to the following circuit.
Yes, you can convert the circuit diagram by replacing the impedance in parallel to the current source even after converting to the Laplace domain( This is because Laplace transform is simply domain transformation for simplification of calculation and has nothing to do with the circuit itself).
But, the way you've transformed the circuit isn't correct(on the L.H.S voltage source, R.H.S. is fine), since the voltages across the L.H.S and R.H.S circuit aren't the same which you can verify yourself.
Alternatively, if you really want to convert to a voltage source, first convert the 3 ohm connected to the ground as two 6 ohm resistors connected in parallel to the ground and then you can use one 6 ohm for converting current to voltage source like you've done above.
Best Answer
There will be more to it than the basic reason, but -
A capacitor's voltage can not be changed instantaneously, but the current into it and out of it can.
If you have a capacitor that you wish to charge to say 100 volts and can charge at up to 1A from the supply available.
If you apply a 100V power supply to the capacitor when it is fully discharged (so at 0V across the capacitor), then as the voltage CANNOT change instantaneously the supply will experience a hard short circuit. 1A (the supply max) will flow into the capacitor and the voltage will rise, but the power supply will in most cases "not be happy.
If you apply a 1A constant current source to the same discharged capacitor, it will again charge at 1A and its voltage will increase, but as the supply is designed to operate in this manner there will be no problem. The capacitor will charge until some preset voltage limit is reached - usually the charging supply in this scenario will be set to 1A constant current with a 100V maximum voltage limit.
Inductors have the opposite constraint - they WILL accept a voltage step but the current they conduct cannot change instantaneously. To "charge" an inductor to 100V across it and 1A flowing in it you could apply a 100V step voltage and the current will increase until the desired 1A current is reached.