Applying electrical equations

Start with what the quantities mean

Electric current, $I$, tells us how much charge, $Q$, passes a point each second: $Q=It$. Potential difference, $V$, tells us how much energy, $E$, is transferred for each coulomb of charge passing between two points: $E=QV$. A potential difference of 1 V means 1 J of energy is transferred per coulomb. Resistance, $R$, describes how strongly a component opposes the current; the relationship used in circuit calculations is $V=IR$.

Power, $P$, is the rate of energy transfer. In a circuit device, $P=IV$: multiplying the energy transferred per coulomb by the coulombs passing each second gives the energy transferred each second. Since $Q=It$, the energy equation can also be written . Using gives the same result. For a resistor, substituting into gives . Choose the equation that uses quantities you know and contains the quantity you need.

Replace series resistors with one equivalent resistance

Resistors in series lie along a single path, so the same current flows through each one. Their individual resistances add:

$R_{\text{total}}=R_1+R_2+\cdots$

This total, or equivalent, resistance would draw the same current from the supply as the original group. Adding a resistor in series increases the opposition to current. The supply potential difference is shared between the resistors, so $V_{\text{supply}}=V_1+V_2+\cdots$. Once you have found the current using , use that same current with to find the potential difference across each resistor.

Link the equations in one circuit

Imagine a 12 V battery connected to a 4 Ω resistor and an 8 Ω resistor in series. Their equivalent resistance is $4+8=12\ \Omega$, so the current everywhere in the loop is $12/12=1\ \text{A}$. The are and ; together they account for the battery's 12 V.

Suppose this steady current flows for 30 s. The charge passing any point is $Q=It=1\times30=30\ \text{C}$. The battery transfers $E=QV=30\times12=360\ \text{J}$. As another check, its transfer is , and . Across the 4 Ω resistor alone, ; across the 8 Ω resistor it is 8 W. These add to the 12 W transferred by the supply.

Match the time and energy units

When power is in watts and time is in seconds, $E=Pt$ gives energy in joules. When power is in kilowatts and time is in hours, it gives energy in kilowatt-hours (kWh). For example, a 0.12 kW device operating for 2 h transfers $0.12\times2=0.24\ \text{kWh}$. A kilowatt-hour is a unit of energy, not power; .

Equations and units

$Q=It$ (C, A, s); $V=IR$ (V, A, Ω); $E=QV$ (J, C, V); (W); (J when is in W and is in s). is energy transferred per second; is energy transferred per coulomb.