Add the individual resistances: $R_{\text{eq}, s} = \sum_{i} R_{i}$

Use the reciprocal sum formula: $\frac{1}{R_{\text{eq}, p}} = \sum_{i} \frac{1}{R_{i}}$

Subtract the voltage drop across the internal resistance from the emf: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$

The total resistance of a group of resistors, treated as a single resistor to simplify circuit analysis.

The potential difference across the terminals of a battery when no current is flowing; the ideal voltage of the battery.

A battery that provides a constant voltage regardless of the current and has zero internal resistance.

An ammeter with zero resistance, so it doesn't affect the current it measures.

A voltmeter with infinite resistance, so it doesn't draw any current from the circuit.

The resistance within a real battery that impacts the voltage it delivers, causing the terminal voltage to be less than the emf when current flows.

It decreases the overall resistance of the circuit, providing more paths for current to flow.

It reduces the terminal voltage of the battery when current is flowing.

The measured current will be lower than the actual current in the circuit.

The measured voltage will be lower than the actual voltage across the component.