Corrected exercises: direct-current circuits (203-SN2-RE)

$Q=I\,\Delta t=(2.0)(30)=60$ C. Number of electrons: $n=\frac{60}{1.6\times 10^{-19}}=3.75\times 10^{20}$, hundreds of billions of billions: current is a river of tiny charges.

$R=\frac{V}{I}=\frac{120}{10}=12$ Ω. Power: $P=VI=1200$ W. Energy: $E=P\,\Delta t=(1200)(180)=2.16\times 10^{5}$ J, that is 216 kJ in three minutes.

Cross-sectional area: $A=\pi r^{2}=\pi(0.50\times 10^{-3})^{2}\approx 7.85\times 10^{-7}$ m². $R=\frac{\rho L}{A}=\frac{(1.7\times 10^{-8})(10)}{7.85\times 10^{-7}}\approx 0.22$ Ω. A small resistance but not zero: over long distances, it is what causes the line losses.

$R=\frac{\rho L}{A}$: doubling $L$ multiplies $R$ by 2; doubling the diameter multiplies $A=\pi r^{2}$ by 4, so it divides $R$ by 4. Net result: $R$ is divided by 2 (about 0.11 Ω). The area, which goes as the square of the radius, weighs more than the length: this is why high-power cables are mostly THICK, not short.

Cross-sectional area: $A=\pi r^{2}=\pi(1.0\times 10^{-3})^{2}=3.14\times 10^{-6}$ m². Current density: $J=\frac{I}{A}=\frac{5.0}{3.14\times 10^{-6}}\approx 1.59\times 10^{6}$ A/m². Drift velocity: $v_{d}=\frac{J}{ne}=\frac{1.59\times 10^{6}}{(8.5\times 10^{28})(1.6\times 10^{-19})}\approx 1.17\times 10^{-4}$ m/s, about 0.12 mm per second.

$t=\frac{3.0}{1.17\times 10^{-4}}\approx 2.6\times 10^{4}$ s, more than SEVEN HOURS. The signal takes $\frac{3.0}{3.0\times 10^{8}}=1.0\times 10^{-8}$ s, ten nanoseconds: a ratio of the order of $10^{12}$. The lamp lights up right away because the electrons at the switch do not have to reach the lamp: the electric field is established throughout the circuit almost instantly and sets the electrons already present everywhere in motion AT THE SAME TIME, including those in the filament. The wire is a pipe already full of water: opening the tap makes water flow at once, without waiting for the first drop to make the trip.

$R=\frac{V^{2}}{P}=\frac{120^{2}}{60}=2.4\times 10^{2}$ Ω. Steady-state current: $I=\frac{P}{V}=\frac{60}{120}=0.50$ A.

$R_{hot}=R_{0}[1+\alpha\,\Delta T]$ with $\Delta T=2500-20=2480$ °C, so $1+\alpha\Delta T=1+(4.5\times 10^{-3})(2480)=12.16$. Hence $R_{0}=\frac{240}{12.16}\approx 19.7$ Ω. At switch-on, the filament is still cold: $I_{0}=\frac{120}{19.7}\approx 6.1$ A, that is 12.2 times the rated current.

During the few tens of milliseconds while the filament heats up, it takes a current twelve times too large, hence a power of the order of $RI^{2}\approx 730$ W instead of 60 W. If a portion of the filament has already been thinned by evaporation of the tungsten, this overload is what melts it: failure therefore happens statistically at switch-on, not after hours of operation. For a metal, resistance rises with $T$ because the lattice ions vibrate more and scatter the electrons more often, while their DENSITY does not change. In a semiconductor, thermal agitation instead tears new carriers from their bonds: $n$ increases exponentially and far outweighs the increased scattering, so the resistance DROPS. Two different mechanisms, two opposite signs for $\alpha$.

Parallel first: $R_{23}=\frac{R_{2}R_{3}}{R_{2}+R_{3}}=\frac{(12)(4)}{16}=3.0$ Ω. Then the series: $R_{eq}=6+3=9.0$ Ω.

$I=\frac{V}{R_{eq}}=\frac{18}{9}=2.0$ A: this current passes through the source and all of $R_{1}$.

$V_{1}=R_{1}I=(6)(2)=12$ V; this leaves $V_{23}=18-12=6.0$ V across the parallel group, the SAME voltage for $R_{2}$ and $R_{3}$. Currents: $I_{2}=\frac{6}{12}=0.50$ A and $I_{3}=\frac{6}{4}=1.5$ A. Junction: $0.50+1.5=2.0$ A, the books balance.

$P_{1}=R_{1}I^{2}=(6)(4)=24$ W; $P_{2}=\frac{V_{23}^{2}}{R_{2}}=\frac{36}{12}=3.0$ W; $P_{3}=\frac{36}{4}=9.0$ W. Sum: $24+3+9=36$ W $=VI=(18)(2)$: the balance holds. $R_{1}$ heats the most, and within the parallel group it is $R_{3}$, the SMALL resistor: at a common voltage, $P=\frac{V^{2}}{R}$ favours small resistors. “Large resistance = large power” is only true in series, where the current is common ($P=RI^{2}$).

The emf is the energy supplied by the battery per unit charge (its “open-circuit voltage”); the terminal voltage is what remains AFTER the drop across the internal resistance: $V=\varepsilon-rI$. The two coincide only when $I=0$: a disconnected battery, or one measured with an ideal voltmeter.

$I=\frac{\varepsilon}{R+r}=\frac{12}{6.0}=2.0$ A. Terminal voltage: $V=\varepsilon-rI=12-(0.5)(2)=11.0$ V (check: $V=RI=(5.5)(2)=11$ V, consistent).

$I_{sc}=\frac{\varepsilon}{r}=\frac{12}{0.5}=24$ A. All the power is dissipated in the battery itself: $P=rI_{sc}^{2}=(0.5)(576)=288$ W, concentrated in a small closed object: violent heating, with a risk of thermal runaway, leakage or explosion. This is the reason for fuses, and for the ban on carrying batteries with unprotected terminals.

The starter motor, a large load in parallel, makes the total current delivered climb: the internal drop $rI$ becomes significant and the terminal voltage $V=\varepsilon-rI$ temporarily collapses. The headlights, connected to those same terminals, receive less voltage and dim. As soon as the engine starts and the current falls back, $V$ recovers: the phenomenon reveals the physical existence of $r$, invisible under normal conditions.

Junction: $I_{3}=I_{1}+I_{2}$. Left loop: $\varepsilon_{1}=R_{1}I_{1}+R_{3}I_{3}$, that is $12=2I_{1}+3I_{3}$. Right loop: $\varepsilon_{2}=R_{2}I_{2}+R_{3}I_{3}$, that is $6=I_{2}+3I_{3}$.

From the loops: $I_{1}=\frac{12-3I_{3}}{2}$ and $I_{2}=6-3I_{3}$. The junction rule gives: $I_{3}=\frac{12-3I_{3}}{2}+6-3I_{3}$. Multiplying by 2: $2I_{3}=12-3I_{3}+12-6I_{3}$, that is $11I_{3}=24$, so $I_{3}=\frac{24}{11}\approx 2.18$ A. Then $I_{1}=\frac{30}{11}\approx 2.73$ A and $I_{2}=-\frac{6}{11}\approx -0.55$ A.

The negative sign of $I_{2}$ means that the actual current flows in the OPPOSITE direction to the assumed one: it does not leave $\varepsilon_{2}$, it enters it through the + terminal. The 6 V source is not delivering current: it is being RECHARGED by the 12 V source, which is stronger. There is no need to redo the calculation: the Kirchhoff formalism has already said everything, you only had to read it.

Right loop: $R_{2}I_{2}+R_{3}I_{3}=(1)\left(-\frac{6}{11}\right)+(3)\left(\frac{24}{11}\right)=\frac{-6+72}{11}=\frac{66}{11}=6.0$ V $=\varepsilon_{2}$: verified. Voltage across $R_{3}$: $V_{3}=R_{3}I_{3}=(3)\left(\frac{24}{11}\right)=\frac{72}{11}\approx 6.5$ V.

In series, the part of the circuit between the two capacitors is isolated: the charge $+Q$ that appears on one plate induces $-Q$ on the neighbouring plate, by conservation of charge of this isolated segment (initially neutral). So $Q_{1}=Q_{2}=Q$. The voltages add: $V=V_{1}+V_{2}=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}$. Dividing by $Q$: $\frac{V}{Q}=\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$.

$C_{eq}=\frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{(6)(3)}{9}=2.0$ μF. Common charge: $Q=C_{eq}V=(2.0\times 10^{-6})(12)=24$ μC. Voltages: $V_{1}=\frac{Q}{C_{1}}=\frac{24}{6}=4.0$ V and $V_{2}=\frac{24}{3}=8.0$ V (sum 12 V: consistent). It is the SMALL capacitor that bears the larger voltage.

$C_{eq}=C_{1}+C_{2}=9.0$ μF, each under 12 V: $Q_{1}=(6.0\times 10^{-6})(12)=72$ μC and $Q_{2}=36$ μC.

Series: $E=\frac{1}{2}C_{eq}V^{2}=\frac{1}{2}(2.0\times 10^{-6})(144)=1.44\times 10^{-4}$ J. Parallel: $E=\frac{1}{2}(9.0\times 10^{-6})(144)=6.48\times 10^{-4}$ J, that is 4.5 times more: the ratio of the energies is that of the equivalent capacitances ($\frac{9}{2}=4.5$), since $V$ is the same. To store energy, connect in parallel.

For resistors, the quantity that adds in series is the OBSTACLE ($V=RI$: the voltage drops stack up). For capacitors, the relation is inverted: $V=\frac{Q}{C}$ puts $C$ in the denominator, so stacking voltages in series adds the $\frac{1}{C}$. In parallel, at a common voltage, resistors add their currents (hence the $\frac{1}{R}$) while capacitors add their charges (hence the $C$). Same circuit logic, inverted quantity.

$C=\frac{\varepsilon_{0}A}{d}=\frac{(8.85\times 10^{-12})(2.0\times 10^{-2})}{5.0\times 10^{-4}}=3.54\times 10^{-10}$ F, that is 354 pF. Charge: $Q=CV=(3.54\times 10^{-10})(12)=4.25\times 10^{-9}$ C. Field: $E=\frac{V}{d}=\frac{12}{5.0\times 10^{-4}}=2.4\times 10^{4}$ V/m. Energy: $U=\frac{1}{2}CV^{2}=\frac{1}{2}(3.54\times 10^{-10})(144)\approx 2.55\times 10^{-8}$ J.

$V_{max}=E_{breakdown}\,d=(3.0\times 10^{6})(5.0\times 10^{-4})=1.5\times 10^{3}$ V. Halving $d$ doubles $C$ but $V_{max}$ falls to 750 V: the field cannot be fooled. The trade-off is structural, since the maximum storable energy $\frac{1}{2}CV_{max}^{2}=\frac{1}{2}\varepsilon_{0}E_{breakdown}^{2}\,(Ad)$ depends only on the VOLUME of dielectric and its strength, not on how it is split between $C$ and $V$. Bringing the plates closer therefore gains no energy: to store more, you need a better insulator or more volume.

The battery imposes $V=12$ V. $C'=\kappa C=1.42\times 10^{-9}$ F, $Q'=\kappa Q=1.70\times 10^{-8}$ C, $U'=\kappa U\approx 1.02\times 10^{-7}$ J: everything is multiplied by 4, except the voltage. The energy comes from the BATTERY, which had to move the extra charge $\Delta Q=1.27\times 10^{-8}$ C through 12 V, supplying $W=\Delta Q\,V\approx 1.53\times 10^{-7}$ J. Yet the capacitor gained only $7.65\times 10^{-8}$ J, exactly HALF. The other half has not vanished: it went into pulling the slab inward and was dissipated in the resistance of the circuit. This factor of 2 is a classic exam question.

With the battery disconnected, the charge is trapped: $Q=4.25\times 10^{-9}$ C stays constant. $C'=1.42\times 10^{-9}$ F, so $V'=\frac{V}{\kappa}=3.0$ V and $U'=\frac{Q^{2}}{2C'}=\frac{U}{\kappa}\approx 6.37\times 10^{-9}$ J. The energy was DIVIDED by 4: it was given up as mechanical work on the slab, pulled inward. If you hold the slab back, you recover this work; if you let go, it accelerates inward and oscillates until friction dissipates the difference.

Energy argument, at constant charge: $U=\frac{Q^{2}}{2C}$ decreases as $C$ increases, and a system evolves spontaneously toward lower energy, so the slab is drawn inward. At constant voltage, the balance in c) shows that the battery supplies twice what the capacitor stores, and the surplus is precisely this pulling work: same conclusion. Microscopically, the field polarizes the dielectric: the molecules orient or deform, and bound charges of opposite sign appear on each face of the slab. These charges create a field that OPPOSES the initial field, which falls from $E$ to $\frac{E}{\kappa}$. For a given charge, a weaker field means a smaller voltage $V=Ed$, hence a larger $C=\frac{Q}{V}$. And since the field is more intense at the edges of the plates, it draws the bound charges toward the strong-field region: hence the pull.

Same current in both resistors: $V_{s}=\varepsilon\frac{R_{2}}{R_{1}+R_{2}}=12\times\frac{4.0}{12.0}=4.0$ V.

The load goes in parallel with $R_{2}$: $R_{2}\parallel R_{L}=\frac{(4)(4)}{8}=2.0$ kΩ. New output: $V_{s}=12\times\frac{2.0}{8.0+2.0}=2.4$ V, that is 40 % below the advertised value. With $R_{L}=400$ kΩ: $R_{2}\parallel R_{L}=\frac{(4)(400)}{404}\approx 3.960$ kΩ and $V_{s}=12\times\frac{3.960}{11.960}\approx 3.97$ V, within 0.7 %. Condition: $R_{L}\gg R_{2}$. A divider delivers the calculated voltage only if almost no current is drawn from it, and this is exactly why a voltmeter must have a very large resistance: measuring already means disturbing.

The circuit is a simple series loop: $I=\frac{\varepsilon}{R+r}$, so $P=RI^{2}=\frac{\varepsilon^{2}R}{(R+r)^{2}}$. Derivative: $\frac{dP}{dR}=\varepsilon^{2}\frac{(R+r)^{2}-R\cdot 2(R+r)}{(R+r)^{4}}=\varepsilon^{2}\frac{r-R}{(R+r)^{3}}$. It vanishes for $R=r$, changing sign from plus to minus: it is indeed a maximum. $P_{max}=\frac{\varepsilon^{2}}{4r}=\frac{144}{16}=9.0$ W.

$P(2.0)=\frac{(144)(2)}{36}=8.0$ W and $P(8.0)=\frac{(144)(8)}{144}=8.0$ W: the SAME value. This is no accident: setting $R=xr$, $P=\frac{\varepsilon^{2}}{r}\frac{x}{(1+x)^{2}}$, an expression unchanged when $x$ is replaced by $\frac{1}{x}$. The curve is therefore symmetric on a logarithmic scale about $R=r$, and it is very flat: being off by a factor of 2 on the load costs only 11 % of the power.

Efficiency: $\eta=\frac{RI^{2}}{(R+r)I^{2}}=\frac{R}{R+r}$. For $R=r$: $\eta=50$ %, half the energy heats the source itself. For $R=36$ Ω: $P=\frac{(144)(36)}{1600}=3.24$ W but $\eta=\frac{36}{40}=90$ %. A distribution grid carries megawatts: losing half of it in the lines would be ruinous and thermally impossible, so it operates far from the maximum, with $R\gg r$, and aims for EFFICIENCY. A radio transmitter or an antenna stage, on the other hand, has a tiny amount of power available and wants to radiate as much of it as possible, even if as much is dissipated in the final stage: it aims for TRANSFER, hence impedance matching at 50 Ω. Maximizing the power delivered and maximizing the efficiency are two different goals, and often opposed.

The shunt must divert everything the galvanometer does not take, under the same voltage: $R_{s}I_{s}=R_{g}I_{g}$ with $I_{s}=I-I_{g}=2.0-0.0010=1.999$ A. Hence $R_{s}=\frac{(50)(0.0010)}{1.999}\approx 2.50\times 10^{-2}$ Ω, about 25 milliohms. Resistance of the whole: $R_{A}=\frac{R_{g}R_{s}}{R_{g}+R_{s}}=\frac{R_{g}I_{g}}{I}=\frac{(50)(0.0010)}{2.0}=2.50\times 10^{-2}$ Ω, practically the shunt alone.

Full scale $I_{g}$ must correspond to 10 V across the whole: $R_{g}+R_{m}=\frac{V}{I_{g}}=\frac{10}{0.0010}=1.0\times 10^{4}$ Ω, so $R_{m}=10000-50=9.95\times 10^{3}$ Ω. Sensitivity: $\frac{10^{4}}{10}=1.0\times 10^{3}$ Ω/V, a mediocre value for modern electronics but typical of moving-needle instruments.

The ammeter we built is worth only 25 mΩ: inserted IN SERIES in a branch, it hardly changes the total resistance, so it hardly changes the current it claims to measure. The voltmeter is worth 10 kΩ: connected IN PARALLEL, it diverts only a trickle of current and hardly disturbs the voltage. Each is designed to be invisible in its own mode of connection. If they are swapped: the voltmeter in series adds 10 kΩ and chokes the circuit, which practically stops working, but nothing is destroyed. The ammeter in parallel across a source, on the other hand, places 25 mΩ across the battery: that is a SHORT CIRCUIT, with a current limited only by the internal resistance. A blown fuse at best, a destroyed instrument at worst. It is the costliest laboratory mistake, and it is never symmetric.

At balance, no current flows through the galvanometer: the two midpoints are at the same potential. Each branch is then a simple voltage divider, carrying a single current. The equality of the midpoint potentials is written $\frac{R_{2}}{R_{1}+R_{2}}=\frac{R_{x}}{R_{3}+R_{x}}$, which reduces to $R_{1}R_{x}=R_{2}R_{3}$, that is $R_{x}=\frac{R_{2}R_{3}}{R_{1}}=\frac{(500)(3200)}{1000}=1.60\times 10^{3}$ Ω.

The condition $R_{1}R_{x}=R_{2}R_{3}$ contains neither the emf nor $R_{g}$: raising the voltage multiplies both branch currents by the same factor and does not shift the balance; as for the galvanometer, it carries ZERO current at balance, so its resistance enters no equation. This is the principle of so-called null methods: you do not measure a quantity, you cancel a difference. The advantage over an ohmmeter is decisive, since an ohmmeter's reading depends on the state of its battery, the calibration of its dial and the resistance of the test leads. Here, the accuracy of $R_{x}$ depends only on that of three standard resistors; the battery only sets the SENSITIVITY, that is, how finely the zero can be detected.