Energy from a battery
- A 9 V battery pushes 5 C of charge through a circuit.
- w = v·q (for constant voltage).
- w = 9 V × 5 C = 45 J.
Current tells you charge is moving; voltage tells you why. Moving charge through a circuit element either costs energy or releases it, and voltage is the exchange rate: the energy transferred per coulomb of charge moved between two points.
One volt (V) is one joule per coulomb. A 9 V battery does 9 joules of work on every coulomb it pushes from its − terminal to its + terminal. This is exactly analogous to gravitational potential energy: lifting a mass to a height stores energy; letting it fall releases it. Voltage is electrical height.
On a schematic, a voltage is labelled with + and − marks. The statement v_AB = 5 V means point A is 5 V above point B. Equivalently, v_BA = −5 V — a negative voltage is not exotic, it just means the polarity is opposite to the labels. Getting comfortable with "negative just means opposite to the arrow/label" now will save you endless grief in nodal analysis (CA-3).
Travelling through an element from − to + is a voltage rise (energy gained by the charge — what happens inside a source). Travelling from + to − is a voltage drop (energy released — what happens in a resistor or an LED). A single element is a rise or a drop depending on which way you traverse it; the physics doesn't change, only your bookkeeping direction.
Moving 20 mC of charge through an element releases 0.1 J. What is the voltage across it?
A phone charger operates at 5 V. How much energy does it give each coulomb, and how much energy is delivered if 3600 C flows during a charge?
If v_AB = 12 V, what is v_BA, and which point is at the higher potential?
Node A is at 3.3 V and node B is at 1.8 V, both measured from ground. What is v_AB? (These are the two logic-supply voltages you'll meet constantly in microelectronics.)