Words for a network
A circuit diagram is a map of connections, not a picture of the apparatus. A junction, also called a node, is a point where three or more wires meet. A branch is the stretch of circuit between two junctions, and one current flows all along it. A loop is any closed path round the network. A wire of negligible resistance is one point electrically, however long it is drawn.
What goes in comes out
In a steady circuit nothing changes with time, so no charge can pile up anywhere. If more current entered a junction than left it, charge would gather there, and its potential would keep rising. So the currents into a junction must equal the currents out of it. This is the first of Kirchhoff's laws.
At any junction of a circuit, the sum of the currents entering equals the sum of the currents leaving. Written with every current taken positive towards the junction, \(\sum I = 0\). The rule expresses the conservation of charge.
In Fig 3.1, \(4 = 1 + 2 + I\), so \(I = 1\,\text{A}\), leaving as drawn. You need not know in advance which way a current flows. Draw an arrow on every branch and apply the rule. If an answer comes out negative, the current flows against your arrow, with the size you found. The algebra is not wrong; it is telling you the direction.
Currents of \(3\,\text{A}\) and \(5\,\text{A}\) enter a junction, and \(2\,\text{A}\) leaves by a third wire. The fourth wire must carry \(8 - 2 = 6\,\text{A}\), and it must carry it away from the junction.
At a junction, \(2\,\text{A}\) and \(3\,\text{A}\) flow in along two wires. You draw the current in a third wire as \(I\) flowing in, and a fourth wire carries \(4\,\text{A}\) out. What is \(I\)?
Back to where you started
Potential is a number attached to each point, as the chapter Electric Potential, Energy and the Dipole showed. Walk once round a closed loop, adding every rise and fall of potential on the way. You arrive back at your starting point, so you arrive back at your starting potential. The changes must add to zero. A charge carried once round the loop gains and loses equal energy.
The algebraic sum of the changes in potential round any closed loop of a circuit is zero, \(\sum \Delta V = 0\). The rule expresses the conservation of energy.
The four signs
The rule is only as good as its signs, and there are four cases. Inside a resistor, current flows from high potential to low. So walking through a resistor along the current, the potential falls: \(\Delta V = -IR\). Walking against the current, it rises: \(+IR\). A cell's positive terminal is higher than its negative one by its emf \(\mathcal E\). Crossing a cell from \(-\) to \(+\) gives \(+\mathcal E\), and from \(+\) to \(-\) gives \(-\mathcal E\).
The sign of a cell does not depend on which way the current flows through it. That is the one place where students lose marks. A cell's terminals keep their order of potential even when a stronger cell pushes current backwards through it. A cell's internal resistance \(r\) is drawn as a separate resistor in its branch and signed like any other resistor.
Here is part of a larger circuit. A current of \(2\,\text{A}\) flows from A to B, first through a \(3\,\Omega\) resistor and then through a \(6\,\text{V}\) cell, entered at its \(+\) terminal. Walk from A to B: \(V_A - 2 \times 3 - 6 = V_B\). So \(V_A - V_B = 12\,\text{V}\). Turn the cell round, so the walk enters at \(-\): \(V_A - 6 + 6 = V_B\), and \(V_A - V_B = 0\).
A point A is at \(10\,\text{V}\). Going from A to B you pass through a \(4\,\Omega\) resistor carrying \(2\,\text{A}\) in your direction. Then you pass through a \(5\,\text{V}\) cell from its negative to its positive terminal. Find the potential of B.
Through the resistor, with the current: \(-2 \times 4 = -8\,\text{V}\). Through the cell, from \(-\) to \(+\): \(+5\,\text{V}\).
\(V_B = 10 - 8 + 5\) \(= 7\,\text{V}\).
Values used: the numbers of the question; the cell is ideal, with no internal resistance.
A single loop holds resistors of \(3\), \(2\), \(1\) and \(4\,\Omega\) and four ideal cells. Going clockwise, you cross the \(30\,\text{V}\) and \(20\,\text{V}\) cells from \(-\) to \(+\), and the \(15\,\text{V}\) and \(10\,\text{V}\) cells from \(+\) to \(-\). What is the clockwise current, in amperes?
One loop, two cells
A loop holds a \(12\,\text{V}\) cell, a \(2\,\Omega\) resistor, a \(3\,\text{V}\) cell and a \(4\,\Omega\) resistor. The \(3\,\text{V}\) cell faces the other way, so it opposes the \(12\,\text{V}\) cell. Guess the current \(I\) clockwise, out of the \(12\,\text{V}\) cell's \(+\) terminal. Walk clockwise: \(+12 - 2I - 3 - 4I = 0\). So \(6I = 9\) and \(I = 1.5\,\text{A}\).
Press play in Fig 3.2. The potential jumps by \(12\,\text{V}\) inside the \(12\,\text{V}\) cell. It falls \(3\,\text{V}\) in the \(2\,\Omega\) resistor and \(3\,\text{V}\) in the opposing cell. The \(4\,\Omega\) resistor takes the last \(6\,\text{V}\), and the dot is back at its starting height. The \(4\,\Omega\) resistor has \(1.5 \times 4 = 6\,\text{V}\) across it.
The current enters the \(3\,\text{V}\) cell at its \(+\) terminal and leaves at its \(-\). Inside it, charge is pushed from high potential to low, against the cell's own push. So the cell is being charged: it takes in energy, \(3 \times 1.5 = 4.5\,\text{W}\), instead of giving it out. A cell whose current enters at \(+\) is always being charged.
Reverse the \(3\,\text{V}\) cell so that it helps. Now \(+12 - 2I + 3 - 4I = 0\), so \(I = 15/6 = 2.5\,\text{A}\), and the \(4\,\Omega\) takes \(10\,\text{V}\). Take two identical cells joined to oppose each other round a loop through a resistor. Then \(\mathcal E - \mathcal E - IR = 0\), so no current flows and the resistor has no voltage across it.
A \(9\,\text{V}\) cell and a \(6\,\text{V}\) cell are joined in one loop through a \(3\,\Omega\) resistor, with their emfs opposing. What happens?
Potentials round a loop
A \(12\,\text{V}\) cell drives current round resistors of \(1\), \(2\) and \(3\,\Omega\). The current is \(12/6 = 2\,\text{A}\). Let S be the cell's \(+\) terminal and P the point after the \(1\,\Omega\). Q comes after the \(2\,\Omega\), and R after the \(3\,\Omega\), at the \(-\) terminal. The circuit alone fixes only differences of potential. To give each point a number, one point must be chosen as zero.
Joining a point to the Earth, earthing it, makes that point zero, as the chapter Conductors in Electrostatics showed. Earth R. Then S is \(12\,\text{V}\). Walking with the current, P is \(12 - 2 = 10\,\text{V}\) and Q is \(10 - 4 = 6\,\text{V}\). R is \(6 - 6 = 0\), as it must be.
Now move the earth wire to Q. Q becomes zero, and every other point drops by \(6\,\text{V}\): S is \(6\,\text{V}\), P is \(4\,\text{V}\) and R is \(-6\,\text{V}\). The current is unchanged. The earth wire carries no current, since it touches the loop at one point only, and the loop equation never mentioned the zero.
Walk once round from S with the current: \(-2 \times 1 - 2 \times 2 - 2 \times 3 + 12\) \(= -2 - 4 - 6 + 12 = 0\). The three drops across the resistors add up to the rise in the cell. This is the loop rule in its plainest form.
In the loop of this section, with \(12\,\text{V}\) and resistors of \(1\), \(2\) and \(3\,\Omega\), the point P is earthed instead. What is the potential of R, in volts?
Why does moving the earth connection from one point of a single loop to another leave the current unchanged?
The earth wire joins the loop at one point only, so no current can flow through it; it has no other path back. Earthing only decides which point is called zero. The loop equation contains differences of potential, the emfs and the IR drops, and none of them depends on where the zero is. So the current is the same, and every potential shifts by the same amount.
In 1845 Gustav Kirchhoff was a student of twenty-one at the University of Königsberg. In a seminar paper on currents in a flat plate, he stated the two rules of this lesson for networks of wires. They reduced any network, however tangled, to a set of equations that could be solved. Kirchhoff later worked with Robert Bunsen on the spectra of the elements, and he gave another famous law, on the radiation of hot bodies.
- Giving a cell's emf a sign from the direction of the current through it.The sign depends only on the terminal you enter by: \(-\) to \(+\) is \(+\mathcal E\), whatever the current does.
- Correcting a negative current by redoing the problem.A negative answer is right. It means the current flows against the arrow you drew, with the size you found.
- Thinking that the earth connection draws current from the loop.An earth wire touching a loop at one point has no return path, so it carries no current. It fixes only the zero.