Exercises — Define voltage (potential difference) and its units

WHAT: Pick the unit. WHY this reasoning: voltage is energy per charge, so its unit must be an energy unit divided by a charge unit. Energy = joule (J), charge = coulomb (C), so voltage = J/C. Answer: (c) joule per coulomb, which we name the volt. (a) is energy alone, (b) is charge alone, (d) is current — none of them is a ratio of energy to charge.

1 joule to move 1 coulomb. That is exactly $1\text{V}=1\frac{\text{J}}{\text{C}}$ — one unit of energy per one unit of charge. On the map (s01), that is the plum arrow having height 1, and one orange coulomb needing 1 J to cross it.

False. Voltage is a difference between two points, written $V_{AB}=V_A-V_B$ (defined above). A single point only gets a number once we pick a ==reference (ground = 0 V)==. "5 V at node X" secretly means "5 V relative to ground." In s01 you cannot draw the plum arrow with only one teal level — you need two levels to have a height difference.

WHAT: find the voltage (the height of the hill in s01) from a known energy and charge. WHY this tool: we're given energy and charge and asked for energy-per-charge — that is $V=W/q$. $V=\frac{W}{q}=\frac{20\text{J}}{4\text{C}}=5\text{V}$

WHAT: find total energy delivered as the charge falls down the 6 V hill (s01, orange arrow). WHY this tool: we know the "hill height" $V$ and the amount of charge $q$; we want total energy, so rearrange to $W=qV$. $W=qV=(3\text{C})(6\text{V})=18\text{J}$

WHAT: find how much charge (how many orange coulombs in s01) it takes to release 30 J across a fixed 5 V hill. WHY this tool: energy and voltage known, charge unknown → $q=W/V$. $q=\frac{W}{V}=\frac{30\text{J}}{5\text{V}}=6\text{C}$

Step 1 — WHAT: find total charge. WHY: current is charge-per-second, so $q=It$. $q=It=(2\text{A})(5\text{s})=10\text{C}$ Step 2 — WHAT: convert charge + voltage into energy (send those 10 coulombs down the 12 V hill in s01). WHY: $W=qV$. $W=qV=(10\text{C})(12\text{V})=120\text{J}$ Check with power: $P=VI=(12)(2)=24\text{W}$, and $W=Pt=24\times 5=120\text{J}$ ✓.

WHAT: find the voltage (hill height) across the resistor from its power and current. WHY this tool: power links voltage and current directly, $P=VI$. Rearrange for $V$. $V=\frac{P}{I}=\frac{18\text{W}}{3\text{A}}=6\text{V}$

Step 1 — WHAT: energy the source hands to the charge, $W=qV=(4)(9)=36\text{J}$. Step 2 — WHAT: useful fraction $=\frac{27}{36}=0.75=75\%$. WHY: voltage tells us energy-per-charge supplied; comparing supplied to useful energy gives efficiency.

WHAT: count charge, then energy, then power for a real charging session. Convert time: $t=2\times 60=120\text{s}$. (a) $q=It=(1.5)(120)=180\text{C}$. (b) $W=qV=(180)(5)=900\text{J}$. (c) $P=VI=(5)(1.5)=7.5\text{W}$. Cross-check: $W=Pt=7.5\times 120=900\text{J}$ ✓. WHY the chain: current+time → charge; charge+voltage → energy; voltage+current → power. Three edges of the toolbox triangle.

WHAT: track the energy of a charge going up then down the s01 hill, using the sign convention $V_{BA}=-V_{AB}$. (a) B→A means climbing the hill (orange charge moving up the slope): $W=q{V}_{AB}=(2)(8)=+16\text{J}$ (you do positive work on the charge). (b) A→B is coming back down: the field does the work, so $W=q{V}_{BA}=q(-V_{AB})=(2)(-8)=-16\text{J}$. (c) Net $=+16+(-16)=0\text{J}$. WHY zero: voltage is like height — a closed loop returns to the same height, so no net energy is stored. This is why voltage is called a potential.

WHAT: find the total charge budget of the battery. WHY $q=W/V$: each coulomb costs the battery $V=1.5$ J to lift, so total charge = total energy ÷ energy-per-coulomb. $q=\frac{W}{V}=\frac{9000\text{J}}{1.5\text{V}}=6000\text{C}$

WHAT: apply $W=q{V}_{AB}$ with a negative charge and read the sign. (a) $W=q{V}_{AB}=(-1\text{C})(+5\text{V})=-5\text{J}$. (b) Negative $W$ means the field does positive work on the electron as it goes $B\to A$ — the electron gains kinetic energy going that way. A positive charge would climb (need $+5$ J) going $B\to A$, but a negative charge does the opposite: it is pulled toward the high-potential point $A$. So on the hill picture, positive charges roll downhill (high $V$ → low $V$) while negative charges "roll uphill" (low $V$ → high $V$). The sign of $q$ flips which way is downhill for that charge.

WHAT: compute a difference, then re-reference it. (a) $V_{XY}=V_X-V_Y=12-3=9\text{V}$ (using the definition $V_{AB}=V_A-V_B$). (b) Shift every reading down by 3 V: X reads $12-3=9\text{V}$, Y reads $3-3=0\text{V}$. New difference $=9-0=9\text{V}$ — unchanged. WHY: the difference is physical; the reference is a bookkeeping choice. Moving ground slides both teal levels in s01 up or down together, so their gap (the plum arrow) is untouched.

WHAT: read the degenerate (flat-hill) case off $V=W/q$. $V=\frac{W}{q}=\frac{0\text{J}}{5\text{C}}=0\text{V}$ Physical meaning: the two points are at the same potential — the two teal levels in s01 collapse to one flat line. No energy is needed to move charge between them, so charge feels no push in either direction. Plenty of charge can sit there, but nothing drives it to flow.

WHAT: test whether $V$ depends on $q$. Diagnosis: doubling $q$ doubles the work $W$, because $W=qV$. But voltage is the ratio $V=W/q$; numerator and denominator both double, so the ratio is unchanged. Correct statement: voltage is a property of the two points (the field), not of the test charge. In s01 the height of the plum arrow doesn't care how many orange coulombs you send up it. Only $W$ scales with $q$; $V$ stays fixed.

WHAT: find the elapsed time by going energy → charge → time. Step 1 — charge: $q=W/V=2000/5=400\text{C}$. Step 2 — time: from $I=q/t$, rearrange $t=q/I=400/0.4=1000\text{s}$. Cross-check: $P=VI=(5)(0.4)=2\text{W}$, and $t=W/P=2000/2=1000\text{s}$ ✓. WHY two routes agree: energy = power × time and energy = charge × voltage are the same relationship viewed through different edges of the triangle.