State Kirchhoff's junction law in words.

State Kirchhoff's loop law in words.

At a junction, \(5.0\,\mathrm{A}\) enters and \(2.0\,\mathrm{A}\) leaves through one branch. What current must leave through the other branch?

A \(9.0\,\mathrm{V}\) battery drives \(2.0\,\Omega\) and \(7.0\,\Omega\) resistors in one series loop. Write the loop equation and find the current.

When traversing a resistor in the same direction as the assumed current, what sign should the potential change have? What sign should it have when traversing against the current?

When traversing an ideal battery from its negative terminal to its positive terminal, what potential change is used in a loop equation?

A loop contains a \(12\,\mathrm{V}\) source, a \(2.0\,\Omega\) resistor, a \(4.0\,\Omega\) resistor, and a \(6.0\,\mathrm{V}\) source opposing the \(12\,\mathrm{V}\) source. Find the current.

A node is connected to \(12\,\mathrm{V}\) through \(3.0\,\Omega\) and to ground through \(6.0\,\Omega\). Use a node equation to find the node voltage.

At a node, \(I_1\) enters while \(I_2\) and \(I_3\) leave. If \(I_1=0.80\,\mathrm{A}\) and \(I_2=0.25\,\mathrm{A}\), find \(I_3\).

Two rectangular meshes share a \(4.0\,\Omega\) resistor. The left mesh has a \(12\,\mathrm{V}\) source and a \(2.0\,\Omega\) outer resistor. The right mesh has a \(6.0\,\mathrm{V}\) source and a \(3.0\,\Omega\) outer resistor. Choose clockwise mesh currents \(I_1\) and \(I_2\), with each source acting as a rise in its clockwise mesh direction. Write the two mesh equations.

Solve the mesh equations from the previous question: \(6I_1-4I_2=12\) and \(-4I_1+7I_2=6\). Also find the current through the shared resistor.

A solved current is \(I=-0.40\,\mathrm{A}\) relative to the arrow you assumed. What is the physical interpretation?

For a connected circuit with \(N\) essential nodes, how many independent node-voltage equations are needed? Why not \(N\)?

Why do Kirchhoff equations also require element laws such as \(V=IR\)?

A single loop contains an unknown source \(\mathcal E\), an opposing \(3.0\,\mathrm{V}\) source, and \(4.0\,\Omega\) plus \(2.0\,\Omega\) in series. The current is \(1.5\,\mathrm{A}\) in the direction of \(\mathcal E\). Find \(\mathcal E\).

A bridge circuit has two series arms across the same battery. The ratio \(R_1/R_2\) on one arm equals \(R_3/R_4\) on the other. A galvanometer connects the two middle nodes. What is the galvanometer current? Explain.

A node at voltage \(V\) is connected to \(10\,\mathrm{V}\) through \(2.0\,\Omega\), to ground through \(5.0\,\Omega\), and to \(4.0\,\mathrm{V}\) through \(3.0\,\Omega\). Write and solve the node equation.

Use Kirchhoff's loop law to explain why a resistor cannot dissipate energy around a closed loop unless some other element supplies energy.

In mesh analysis, two mesh currents pass through the same resistor. When should the resistor current be written as \(I_1-I_2\), and when should it be written as \(I_1+I_2\)?

A node at voltage \(V\) is connected to \(18\,\mathrm{V}\) through \(6.0\,\Omega\), to ground through \(3.0\,\Omega\), and to \(-6.0\,\mathrm{V}\) through \(6.0\,\Omega\). Use Kirchhoff's junction law to find \(V\) and the three branch currents.