The picture in your head
Charge is the stuff that moves. Current is how quickly it moves. Voltage is the energy each coulomb is given (or loses). Resistance is the traffic. Keep those four words separate and the equations write themselves.
2.14–2.16, 2.20–2.21
Charge, current, electrons and voltage
Current is the rate of flow of charge (2.14).
$$Q = I \times t$$$Q$ in coulombs, $I$ in amperes, $t$ in seconds. A current of $2.0\,\text{A}$ for $30\,\text{s}$ moves $60\,\text{C}$.
In a metal, the moving charges are negatively charged electrons (2.16). They drift from the negative terminal towards the positive. Conventional current is drawn the other way — positive to negative — because that is the historical arrow. In the exam, “current” means conventional current unless they ask about electrons.
Voltage is the energy transferred per unit charge (2.20). The volt is a joule per coulomb.
$$E = Q \times V$$If $4.0\,\text{C}$ of charge passes through a $12\,\text{V}$ lamp, $E = 48\,\text{J}$ is transferred (mostly to heat and light).
Watch
Current, potential difference and resistance
2.10, 2.13
Ohm’s law
$$V = I \times R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}$$Increase $R$ (same $V$) and $I$ falls (2.10). Increase $V$ (same $R$) and $I$ rises. That is all “qualitative effect of changing resistance” wants.
A $12\,\text{V}$ cell and a $6.0\,\Omega$ resistor: $I = 12/6.0 = 2.0\,\text{A}$. Add a second $6.0\,\Omega$ in series and $R=12\,\Omega$, so $I=1.0\,\text{A}$ — current halves because total resistance doubled.
2.7, 2.8, 2.17–2.19
Series and parallel — learn the rules as a table
| Series (one loop) | Parallel (junctions) | |
|---|---|---|
| Current | Same everywhere. Not “used up”. | Splits at a junction; the currents add back to the total (conserved). |
| Voltage | Shared. $V_1 + V_2 = V_{\text{supply}}$ | Same across each branch as the supply (2.18). |
| Resistance | $R_{\text{total}} = R_1 + R_2$ (2.19) | Total $R$ is less than the smallest branch (qualitative only). |
| If one lamp fails | The loop is broken — all go out. | The other branch still works. That is why house lights are in parallel (2.7). |
Domestic lighting is parallel so each lamp gets the full mains voltage and can be switched independently. Christmas lights that all die when one bulb fails are (old-style) series.
Two resistors in series — the 2.19 calculation
$R_1 = 4.0\,\Omega$, $R_2 = 8.0\,\Omega$, supply $V = 12\,\text{V}$.
$$R_{\text{T}} = 4.0 + 8.0 = 12\,\Omega$$ $$I = 12/12 = 1.0\,\text{A} \quad \text{(same through both)}$$ $$V_1 = IR_1 = 4.0\,\text{V},\quad V_2 = IR_2 = 8.0\,\text{V}$$Check: $4.0 + 8.0 = 12\,\text{V}$. The larger resistor takes the larger share of the voltage.
Watch
Series circuits and Ohm’s law calculations
Check you can
A $9.0\,\text{V}$ battery, $R_1=3.0\,\Omega$ and $R_2=6.0\,\Omega$ in series. Find $R_{\text{T}}$, $I$, $V_1$ and $V_2$.
$R_{\text{T}}=9.0\,\Omega$, $I=1.0\,\text{A}$, $V_1=3.0\,\text{V}$, $V_2=6.0\,\text{V}$.