Electronic – Can you help me find the voltage drop for R3 and R4 on the following circuit

First off, mesh analysis would be simpler because you would only have one unknown (the current in the right mesh).

As you know, in nodal analysis you are summing currents at the nodes. Current into the nodes positive and current out of the nodes negative. The way you have defined the problem you should have 3 equations and three unknowns. Your first equations seems to do this correctly for node B. It kind of falls apart after that. For example, node A equation should just be .005 = (Va-Vb)/3000. You already know that Vc = 16 volts.

So simplifying the node A equation gives you what Andy aka said, then just substitute it back into your node B equation.

To me, Kirchoff's Laws are scientific wording of what should be common-sense observations of electric circuits. Unfortunately, common sense isn't as common as we might like.

KVL says that the total of the voltage drops in a circuit must equal the supplied voltage. If that was not true, we would have some voltage across a wire which would result in approximately infinite current in that wire - but KCL insists that the current is the same at all points in a simple series circuit, so we can't have a huge current at one point in the circuit.

If you connect a device that normally requires 4 volts across a 6 volt battery, sufficient current will flow to make the resulting circuit comply with KVL. The voltage across the "4 volt device" will rise, and the output voltage of the battery will fall (due to a voltage drop across the internal resistance of the battery) such that the voltage across the battery and the device are equal. This may result in the destruction of the 4 volt device, if it cannot withstand the extra voltage and current.

Regarding your second point: the voltage drop across a resistor will depend on its resistance, and on the current passing through it, in accordance with Ohm's Law.

For your example of a 10 Ohm resistor across a 30 volt supply, the current through the resistor will be 3 amps, and the voltage across the resistor will be 30 volts. If you add a second 10 Ohm resistor in series, the load on the power supply is now 20 Ohms, so, by Ohm's Law, the current through the resistors will be 1.5 amps, and there will be 15 volts dropped across each resistor, for a total voltage drop of 30 volts.

If you put the two 10 Ohm resistors in parallel across the 30 volt supply, each resistor will now see 30 volts, and will each pass 3 Amps, so the supply will have to supply 6 Amps.