Circuits

Electromotive Force

To maintain a constant current in a closed circuit, electrical energy must be supplied. The source of the energy is commonly called the electromotive force (emf) $\varepsilon$ :

\varepsilon = \frac{dW}{dq}.

The unit of electromotive force is the volt ( $V$ ).

Assume a circuit with battery with voltage equal to $\varepsilon$ and a resistor with resistance $R$ . Assume the battery has no internal resistance. To drive the current around the circuit, the battery undergoes a discharging process:

The electrostatic force is conservative, meaning the work done to move charge $q$ around the circuit is zero:

W = -q \oint_C \boldsymbol{E} \cdot d\boldsymbol{s} = 0.

Let $a$ in the above image be the starting point. When crossing the battery, the potential increases by $\varepsilon$ and when crossing the resistor, the potential decreases by $IR$ . Assuming the wire carries no resistance, upon completing the loop the net potential difference is zero:

\varepsilon - IR = 0 \implies I = \frac{\varepsilon}{R}.

A real battery always has some internal resistance $r$ :

The voltage across the battery terminals are:

U = \varepsilon - Ir.

Again, the net potential difference around a closed loop is zero:

\begin{align*} U - IR &= \varepsilon - Ir - IR \\ &= \varepsilon - I (r + R) = 0, \\ I &= \frac{\varepsilon}{r + R}. \end{align*}

For a source with emf $\varepsilon$ , the power is:

P = I \varepsilon = I^2 (r + R).

Total Resistance

Resistors with resistance $R_i$ are connected in series to a battery with voltage $U$ . By current conservation, the same current $I$ is flowing through each resistor:

After crossing a resistor, the voltage drops by $U_i = I R_i$ . The total voltage drop $U$ is the sum of the individual voltage drops:

U = \sum_i U_i = \sum_i I R_i = I \sum_i R_i.

The resistors can be replaced by a resistor with resistance $R$ with the identical voltage drop $U = IR$ :

\begin{align*} U = IR &= I \sum_i R_i, \\ R &= \sum_i R_i. \end{align*}

Now, consider resistors with resistance $R_i$ connected in parallel to a battery with voltage $U$ . By current conservation, the current $I$ that passes through the battery is divided into currents $I_i$ that pass through the resistors:

Each resistor has a voltage $U_i = I_i R_i$ , but the voltage across all the resistors is the same:

\begin{align*} U &= U_i = I_i R_i, \\ I &= \sum_i I_i = \sum_i \frac{U}{R_i}. \end{align*}

The resistors may be replaced by a resistor with resistance $R = \frac{U}{I}$ :

\begin{align*} \frac{U}{R} &= \sum_i \frac{U}{R_i}, \\ \frac{1}{R} &= \sum_i \frac{1}{R_i}. \end{align*}

Total Resistance Calculator

Parallel connection

R = 0\ \Omega

R_1 =

Kirchhoff's Circuit Rules

Junction Rule

At any point where there is a junction between various current carrying branches, by current conservation the sum of the ingoing currents must equal the sum of the outgoing currents:

\sum I_{\textrm{in}} = \sum I_{\textrm{out}}.

Consider an example:

The currents are related:

I_1 = I_2 + I_3.

Loop Rule

The sum of voltage drops $U$ across any circuit elements that form a closed loop is zero:

\sum_{\textrm{closed loop}} U = 0.

The rules for determining $U$ across a resistor and a battery with a chosen travel direction are as follows:

note: the choice of travel direction is arbitrary.

As an example, consider a voltage source $\varphi_{in}$ connected to two resistors $R_1$ and $R_2$ :

The voltage difference $\varphi_{\textrm{out}}$ across resistor $R_2$ will be less than $\varphi_{\textrm{in}}$ . This circuit is called a voltage divider. From the loop rule:

\varphi_{\textrm{in}} - I R_1 - I R_2 = 0.

So the current is given by:

I = \frac{\varphi_{\textrm{in}}}{R_1 + R_2}.

The voltage difference $\varphi_{\textrm{out}}$ across resistor $R_2$ is equal to:

\varphi_{\textrm{out}} = I R_2 = \frac{R_2}{R_1 + R_2} \varphi_{\textrm{in}}.

The ratio of the voltages characterizes the voltage divider and is given by the resistances:

\frac{\varphi_{\textrm{out}}}{\varphi_{\textrm{in}}} = \frac{R_2}{R_1 + R_2}.