One leg of the 2Ω resistor is at 16V with respect to the -ve terminal of the 16V source.
The other leg is at 10V.

Hence V1 = 16V - 10V = 6V
I1 = 6V / 2Ω = 3A

What about this you do not understand?

 

Is it possible that the conceptual difficulty here arises from the standard conventions used for schematics?

When a current direction is specified for an unknown current, it does not mean that the current actually flows in that direction, just that it is measured in that direction. Similarly when an unknown voltage is specified with a given polarity, the voltmeter is connected with that polarity. However, when a voltage source is specified with a given polarity or current source with a given direction, then an ideal source will ensure that is the actual polarity or direction. (Just don't put voltage sources in parallel or current sources in series.)

Also, don't try to build this circuit in the lab. Laboratory sources are typically built to source voltage or current, not absorb it like theoretical sources can.

 

Ok, but the voltage there can't be 10V is my point. Electrical components don't work like "ok there's 10V there so the 6V extra from that other supply get dropped across R1". The Current flows won't work out even if that were true. People keep posting numbers, but they only work in specific sections of the circuit and not across the whole. Please detail the math of the components if you see an answer to the problems.

Hello,

That may be why you are having trouble figuring this out. The voltage across the 4 and 6 Ohm resistors is 10 volts for sure because there is a voltage source right across those two that is 10 volts. The other source has no influence on the 10v source because in theory you can not change the voltae of an ideal voltage source. So in this circuit the 10v source may actually be sinking current. If it was a battery it would be charging not discharging.

So with that in mind, see if you can solve it now.

The fact that the 10v source is having power delivered to it rather than delivering power to the circuit may seem unusual to you but this happens a lot in exam type questions, and in real life in battery charging systems all the time. There are other times when a source may be receiving power for short time periods even though supplying power other times. Even the line voltage (120vac) can receive power rather than deliver in various applications. The companies that supply power to your home do it all the time, and in fact that is their business, to supply power to the line.

 

When there is sufficient sun, my electric meter runs backwards and my neighbors are buying power from me instead of from the electric company. Another example of current flowing into a voltage source.

Bob