D.C. Circuits: Question 1

Worked solution

Step 1: Recall Kirchhoff’s first law

Kirchhoff’s first law is a consequence of the conservation of charge: the total current flowing into a junction must equal the total current flowing out of it, since charge cannot build up or disappear at a junction.

Step 2: Add the currents flowing into the junction

Two currents flow into the junction: $5.0\text{ A} + 3.0\text{ A} = 8.0\text{ A}$

Check by adding in the reverse order: $3.0+5.0=8.0\text{ A}$, same total, confirmed.

Step 3: Use the known outgoing current to find $I$

The total current out must also equal $8.0\text{ A}$. One outgoing wire already carries $6.0\text{ A}$, so: $6.0\text{ A} + I = 8.0\text{ A}$ $I = 8.0 - 6.0 = 2.0\text{ A}$

Check: substituting back, $6.0+2.0=8.0\text{ A}$, which matches the total current into the junction. Confirmed.

Step 4: Why the other options are wrong

• B ($4.0\text{ A}$): this comes from splitting the total incoming current ($8.0\text{ A}$) equally between the two outgoing wires, ignoring that one of them is already stated to carry $6.0\text{ A}$.
• C ($8.0\text{ A}$): this treats the fourth wire as if it alone carries all the current entering the junction, forgetting that $6.0\text{ A}$ already leaves along the third wire.
• D ($14.0\text{ A}$): this simply adds all three given current values ($5.0+3.0+6.0$) without separating “in” from “out”, which is not what Kirchhoff’s first law requires.

Final answer

• Total current in $=$ total current out: $8.0\text{ A}=6.0\text{ A}+I$, so $I=\boxed{2.0}\text{ A}$, option A.