Even for a simple circuit with all the electrical elements set up in series, calculation of the amperage, or electrical current, can be complex. If the only element is a resistor, the familiar formula V=IR applies. However, the formulas get increasingly complicated as you add capacitors and inductors. Capacitors slow the current down since they form a gap in the circuit. Inductors slow the current down because their magnetic field opposes the electromotive force driving the current. Oscillating the electromotive force further complicates the equations.

## Direct Current

### Step 1

Calculate the current in a direct-current circuit with resistors in series by summing the resistances. Denote the sum of the resistance with the letter \"R\". Then the current through the circuit is I = V/R, where \"V\" is the electromotive force (emf), in volts, provided by the DC source.

### Step 2

Account for a capacitor added in series with the formula I=(V/R)*exp[-t/RC]. \"V\" and \"R\" are as defined in Step 1. \"C\" is the capacitance of the capacitor, and \"t\" is the time after the closing, or completion, of the circuit. If \"V\" is in volts, \"R\" is in ohms, \"C\" is in Farads and \"t\" is in seconds, then \"I\" is in amps. Here, the asterisk indicates multiplication and \"exp[ ]\" indicates the amount in brackets is the exponent of the number “e,” which equals roughly 2.718. Note that as \"t\" gets large, charge builds up on the capacitor and current slows down, ultimately approaching zero.

### Step 3

Account for inductors in series instead of a capacitor with the formula I==(V/R)*{1-exp[-tR/L]}, where \"L\" is the sum of the inductances of the inductors. Note that the inductors’ opposition to the main voltage decreases with time, and \"I\" converges to \"V/R\". If \"L\" is in Henries, \"I\" is in amps.

## Alternating Current

### Step 1

Account for an alternating electromotive force (emf) by first summing the resistance of the resistors in series and denoting it with the letter \"R\".

### Step 2

Account for a capacitor in series in the circuit by calculating the capacitive reactance, which is 1/(?t), where the electromotive force drives the current with a frequency \"?\". Denote the capacitive reactance by \"Xc\". If there is no capacitor, set \"Xc\" to zero.

### Step 3

Account for all inductors in series by summing their inductance, in Henries, and denoting the sum with the letter \"L\". Then calculate the inductive reactance with the formula \"?L\". Denote it by \"Xl\". If there is no inductor, set \"Xl\" to zero.

### Step 4

Calculate the impedance, which is the square root of R-squared plus the square of the difference of the reactances you found in Step 2 and 3. In other words, the impedance, \"Z\", is the square root of R^2 + (Xl-Xc)^2.

### Step 5

Calculate the maximum value of the current by dividing the maximum value of the emf, denoted \"V\", by the impedance. So I=V/Z.

### Step 6

Solve for the angular separation, or phase constant, between the current and emf peaks by taking the arctangent of (Xl-Xc)/R. Denote it by \"?\". For example, if the emf oscillation is V*sin ?t, then the current oscillation is I*sin(?t-?).