more current needed in an electrical circuit
You can't cheat physics.  You can convert between different voltage and current tradeoffs, but you can't make more power out than in.  Power is Volts x Amps, so whatever Volts x Amps you convert to can't exceed the Volts x Amps you have coming into the converter.  Actually the output will be a bit less due to inevitable loss.  No converter is 100% efficient.

You have 2 V and 1 A available.  (2 V)(1 A) = 2 Watts.  That is your input power, which means it's also the upper limit on your output power.  Let's say you make a converter that is 80% efficient.  80% of 2 W is 1.6 W, which is how much your converter can put out.  That could be, for example, in the form of 10 V and 160 mA, 5 V and 320 mA, etc.

To get 30 A out, the voltage would only be (1.6 W)/(30 A) = 53 mV.  In practice, such a large conversion ratio (30:1 in this case) and very low output voltage will make the converter very inefficient.  You're not going to convert 2 V and 1 A to 30 A at whatever voltage with 80% efficiency.

By using storage, like a battery, you can have temporary peak output power exceed the input power.  However, in the long run, you're still limited by the same laws of physics.  You could, for example, make a circuit that charges a 12 V car battery from the 2 V at 1 A input.  Let's say you can do that at 80% efficiency, so the current into the car battery would be (1.6 W)/(12 V) = 130 mA.  If you draw 30 A from the car battery for a short time, it would take (30 A)/(130 mA) = 230 times longer to recharge the battery, and that's not taking into account the efficiency of the battery itself.  For example, 30 A for 1 minute would take 3 hours 51 minutes of charging time to get back to where you were.  Taking the battery efficiency into account, it's probably closer to 4&frac12; to 5 hours in reality.