Electric Circuits

Current is how much charge passes a point each second: $I = \dfrac{q}{t}$, measured in amperes (coulombs per second). The battery's push is the voltage, a separate quantity.

Conventional current is defined as the flow of positive charge: out of the $+$ terminal, through the circuit, back into the $-$ terminal. (The electrons actually drift the opposite way, which is why the two are often confused.)

Ohm's law, $V = IR$, means that for a fixed resistance the current rises in direct proportion to the voltage. A graph of $I$ against $V$ is then a straight line through the origin.

All the current that passes through the component must also pass through the ammeter, so it goes in series. A voltmeter, which measures the potential difference across a component, is connected in parallel with it.

Charge is not used up – it only carries energy. In a single loop there is nowhere for it to go, so the same current flows through every component. (What the components use up is energy, seen as a voltage drop, not current.)

Both ends of each parallel branch connect to the same two points, so each branch feels the full battery voltage. (It is the current that divides between parallel branches, not the voltage.)

$I = \dfrac{q}{t} = \dfrac{12}{4.0} = 3.0\text{ A}$.

$I = \dfrac{V}{R} = \dfrac{9.0}{3.0} = 3.0\text{ A}$.

Charge cannot pile up at a junction, so the current leaving equals the total current arriving: $2.0 + 3.0 = 5.0\text{ A}$. This is Kirchhoff's current rule – a statement of conservation of charge.

In series the resistances add: $R_\text{total} = 4.0 + 8.0 = 12\ \Omega$. Then $I = \dfrac{V}{R} = \dfrac{12}{12} = 1.0\text{ A}$ – the same current through both resistors.

$\dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}$, so $R = 2.0\ \Omega$. The combined resistance of a parallel pair is always less than either resistor.

$P = VI = (240)(0.50) = 120\text{ W}$.

Rearranging, $R = \dfrac{V^2}{P} = \dfrac{240^2}{60} = \dfrac{57\,600}{60} = 960\ \Omega$.

Resistance is $R = \dfrac{\rho L}{A}$, directly proportional to length. Doubling the length doubles the resistance – a longer wire opposes the flow of charge more.

First the charge: $q = It = (0.50)(60) = 30\text{ C}$. Then the number of electrons: $n = \dfrac{q}{e} = \dfrac{30}{1.6\times10^{-19}} \approx 1.9\times10^{20}$.

First combine the parallel pair: $\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$, so $R_p = 2.0\ \Omega$. Then add the series resistor: $R_\text{total} = 4.0 + 2.0 = 6.0\ \Omega$.

The current is $I = \dfrac{9.0}{6.0 + 3.0} = 1.0\text{ A}$. The voltage across the $6.0\ \Omega$ is then $V = IR = (1.0)(6.0) = 6.0\text{ V}$ – the larger resistor takes the larger share (two-thirds) of the supply.

Energy used $= 2.0\text{ kW} \times 3.0\text{ h} = 6.0\text{ kWh}$. Cost $= 6.0 \times \$0.30 = \$1.80$.

A metal at constant temperature is ohmic – a straight line (A). A filament globe gets hotter as the current increases, so its resistance rises and the $I$-$V$ line bends over, curving away from a straight line (B). It does not obey Ohm's law.

Each parallel branch still gets the full battery voltage no matter what the other branch does. Removing one globe just opens its own branch; the remaining globe keeps the same voltage and current, so it glows exactly as before.

Adding a globe in series raises the total resistance, so the current from the battery drops. With less current, each globe dissipates less power ($P = I^2R$) and glows more dimly. (This is why series fairy-lights dim as you add more.)

The parallel pair is $\dfrac{4}{2} = 2.0\ \Omega$, so the total is $2.0 + 2.0 = 4.0\ \Omega$ and the battery drives $I = \dfrac{24}{4.0} = 6.0\text{ A}$. This $6.0\text{ A}$ splits equally between the two identical $4.0\ \Omega$ branches, so each carries $3.0\text{ A}$.

Doubling the length multiplies $R$ by $2$. Doubling the diameter multiplies the area by $2^2 = 4$, which divides $R$ by $4$. Overall $R \to \dfrac{2}{4}R = \dfrac{1}{2}R$ – it halves.

In series both carry the same current, so $P = I^2R$ is proportional to $R$. The $6.0\ \Omega$ resistor therefore dissipates twice the power of the $3.0\ \Omega$. (The smaller resistor wins only in a parallel arrangement, where $P = \tfrac{V^2}{R}$.)

At its rated voltage, $R = \dfrac{V^2}{P}$, so the $40\text{ W}$ globe actually has the higher resistance and the $60\text{ W}$ globe the lower. In series they carry the same current, and $P = I^2R$, so the higher-resistance $40\text{ W}$ globe dissipates more power and glows brighter – the opposite of what most people expect.