Kirchhoff's laws and loops

Define these three elements of circuit topology in your own words: (a) node, (b) branch, and (c) loop. Give an example of each by describing a simple circuit with two batteries and three resistors.

Match each concept to its definition: (1) node, (2) branch, (3) loop, with: (a) section of a circuit between two consecutive nodes, through which the same current flows; (b) closed path starting and ending at the same point without passing twice through the same branch; (c) point in the circuit where three or more conductors join.

The rules for series and parallel circuits (current is the same in series, voltage is the same in parallel) are actually special cases of Kirchhoff's laws. Explain which law each corresponds to and why Kirchhoff's laws are useful for circuits that are neither series nor parallel.

$5\ \text{A}$ arrive at a node through a cable and $2\ \text{A}$ leave through another branch. Applying Kirchhoff's current law ($I_\text{entran} = I_\text{salen}$), calculate what current leaves through the third branch of the node and explain your reasoning.

Two cables with $4\ \text{A}$ and $2\ \text{A}$ arrive at a node, and $5\ \text{A}$ leave through a third one. (a) Calculate the magnitude and direction (entering or leaving) of the current through the fourth cable. (b) Check the result by writing $\sum I = 0$ with signs (positive for entering, negative for leaving).

Between two nodes there are two parallel branches: through the first, of resistance $2\ \Omega$, $6\ \text{A}$ flow. (a) Applying Kirchhoff's current law, calculate the current in the second branch if $9\ \text{A}$ arrive through the common cable. (b) Calculate the resistance of that second branch knowing that both withstand the same voltage. (c) Explain using Kirchhoff's current law and Ohm's law why the branch with smaller resistance carries more current.

State Kirchhoff's voltage law (or Kirchhoff's second law) in your own words: when travelling completely around any closed path of a circuit, the sum of voltage gains and drops is equal to $\sum V = 0$. Explain it using the analogy of a circular walk on a mountain (you go up and down, but return to the same elevation).

In a loop with a $12\ \text{V}$ battery and two resistors, voltage drops of $5\ \text{V}$ and $7\ \text{V}$ are measured. (a) Verify that Kirchhoff's voltage law is satisfied by writing the signed sum. (b) Reason what would happen to the loop current if the measured drops added up to less than the battery voltage.

A current of $2\ \text{A}$ flows through a branch formed by a battery of $12\ \text{V}$ (traversed from its negative to its positive terminal) and a resistor of $3\ \Omega$. Calculate the potential difference between the ends of the branch by traversing it in the direction of the current (rises $12\ \text{V}$ in the battery and drops $I \cdot R$ in the resistor) and explain each term of your calculation.

The circuit in the diagram has two batteries and three resistors distributed over two loops, with the branch currents labelled, as shown in the diagram.

(a) Write the upper node equation relating the three branch currents ($I_1$, $I_2$ and $I_3$). (b) Write the equation for the left loop (the one containing the battery of $12\ \text{V}$, $R_1$ and $R_3$) by applying $\sum V = 0$. (c) Write the equation for the right loop. It is not necessary to solve the system: only set it up.

A classmate claims to have solved a two-loop circuit (battery of $12\ \text{V}$ with $R_1 = 3\ \Omega$, battery of $8\ \text{V}$ with $R_2 = 2\ \Omega$, and $R_3 = 2\ \Omega$ in the common branch) and says that the currents are $I_1 = 2\ \text{A}$, $I_2 = 1\ \text{A}$ and $I_3 = 3\ \text{A}$. Check whether their solution is correct by verifying, with the given numbers: (a) Kirchhoff's current law at the node and (b) Kirchhoff's voltage law in the two loops.

A two-loop circuit has a battery of $14\ \text{V}$ in series with $R_1 = 2\ \Omega$ in the left branch, a battery of $9\ \text{V}$ in series with $R_2 = 1\ \Omega$ in the right branch, and $R_3 = 2\ \Omega$ in the common branch. (a) Set up the system of Kirchhoff equations with the branch currents $I_1$ and $I_2$ (the current in the common branch is $I_1 + I_2$). (b) Solve it and give the three currents with their units. (c) Check the solution by verifying the node and one of the loops.

The circuit in the diagram has two batteries in the side branches and a resistor in the common branch, with the current of that branch marked with a question mark, as shown in the diagram.

(a) Set up the system of Kirchhoff equations for the circuit. (b) Solve it and calculate the three branch currents, in particular that of the common branch marked in the diagram. (c) Check the result using Kirchhoff's current law at the node.