A minus sign is an answer

A junction is a node where three or more branches meet. It has no capacity to store charge, so whatever arrives per second must leave per second:

$$\sum I_{\text{in}} = \sum I_{\text{out}}, \qquad \text{equivalently} \qquad \sum_{\text{node}} I = 0 \ \text{with currents signed by direction.}$$

What is conserved here is charge per unit time, not energy per unit charge. Voltage is what gets spent crossing elements; current is what is bookkept at nodes. Two consequences follow immediately.

1. Series elements carry identical current, because the node between them has only two connections and there is nowhere else for charge to go.
2. A parallel section's feed current equals the SUM of its branch currents. If a $2\ \Omega$ branch carries $6$ A and a $3\ \Omega$ branch carries $4$ A, the wire feeding them carries $10$ A, and the section has $12$ V across it.

The rule holds instant by instant, even during a transient, which is why it still applies while a capacitor is charging.

Before any equation is written, draw an arrow on every branch and label it $I_1$, $I_2$, $I_3$. The arrows do not have to be right. They only have to be fixed, because every junction and loop equation will be written against them.

When the system is solved, a negative value carries exactly one message: the true current in that branch runs opposite to the arrow you drew, and its magnitude is the number you got. $I_2 = -0.40$ A means $0.40$ A flowing the other way, and the problem is done.

1. Do not redraw and restart. Reversing an arrow after the equations are written and re-solving is extra work that reaches the same physics.
2. Do not flip one sign by hand. Changing $I_2$'s sign in the junction equation while the loop equations still use the old convention corrupts the whole system, and the corrupted system usually still has a solution, so nothing announces the error.
3. Carry the signed value forward. Substituted consistently, $-0.40$ A gives correct potential differences everywhere. Where a magnitude is wanted, such as $P = I^2R$, the square removes the sign on its own.

Flip an arrow only before writing any equation, never after.

A network with $B$ branches has $B$ unknown currents, so it needs $B$ independent equations. They come from the two rules in a fixed proportion:

1. Junction equations: a circuit with $N$ nodes gives $N - 1$ independent ones. The last node's equation is the sum of all the others and adds nothing.
2. Loop equations: the remaining $B - (N - 1)$ come from independent loops, meaning each new loop must include at least one branch no earlier loop used.

A two-node, three-branch circuit therefore needs $1$ junction equation and $2$ loop equations. Writing both node equations and only one loop equation gives three equations that look complete and are not: two of them are the same statement, and the system will not close.

Solve, then check. Substituting the solved currents back into a loop you did not use is a genuine test, and it catches sign errors that the original equations were built around.

A meter has to sample the quantity it reports without changing it, and the two meters do that in opposite ways.

1. Ammeter. It counts the charge passing THROUGH itself, so the branch must be broken and the meter inserted in line. Ideally it has zero resistance, so adding it in series changes nothing.
2. Voltmeter. It compares the potential at two points, so it bridges ACROSS the element. Ideally it has infinite resistance, so adding it in parallel draws no current and changes nothing.

Swapped, each meter destroys the circuit it was supposed to observe. An ammeter clipped across a resistor is a near-zero-resistance path in parallel with it: the resistor is short-circuited, the current bypasses it, and the reading describes the broken circuit. A voltmeter inserted in series is a near-infinite resistance in the branch: the current stops, and the meter reads almost the whole source voltage while the element it was pointed at reads nearly zero.

Real meters are close to those ideals rather than at them, and the residual error goes the way the model predicts. A real ammeter's small resistance lowers the branch current it is measuring, and it then reports that lowered value correctly. A real voltmeter's large but finite resistance loads a divider, pulling the measured node toward the value the parallel combination gives. In both cases the meter is telling the truth about a circuit you slightly changed.