Unit 5

Current, Resistance & DC Circuits

Units 1–4 were electrostatics: charges sitting still, fields inside conductors equal to zero. This unit lets the charges move. Keep a field inside a wire and charge flows — a current. The wire resists, turning electrical energy into heat, and a battery keeps the field going. From those three ideas come Ohm's law, power, Kirchhoff's two rules, and the RC circuit — the first circuit whose behaviour changes in time, and the reason every digital signal has a rise time. This is the day-one toolkit of an electrical engineer. The lab puts an RC circuit on a virtual oscilloscope so you can watch the exponential happen.

1 · Current and drift velocity

Picture first. Inside a copper wire, about one electron per atom is free to wander. With no field they whizz around at roughly \(10^6\) m/s in random directions — like a cloud of gnats — and the net flow is zero. Apply a field \(\mathbf{E}\) along the wire and each electron gets a tiny push between collisions. The cloud as a whole now creeps slowly opposite to \(\mathbf{E}\) (electrons are negative). That slow collective creep is the drift velocity \(v_d\), and it is what carries the current.

Each electron zig-zags between collisions with the lattice; the field only biases the zig-zag. Electrons drift left, opposite to \(\mathbf{E}\); the conventional current — defined as the direction positive charge would move — points right.

\[ I = \frac{dQ}{dt}\qquad [\text{A} = \text{C/s}] \] Current \(I\) is the charge \(dQ\) crossing a cross-section per time \(dt\). One ampere is one coulomb per second, about \(6.24\times10^{18}\) electrons per second.

Count how much charge crosses area \(A\) in time \(dt\). Every carrier within a distance \(v_d\,dt\) of the cross-section gets through, so the charge is (carriers per volume) × (volume) × (charge each) \(= n\,(A v_d\,dt)\,q\):

\[ I = nqv_dA,\qquad \mathbf{J} = nq\mathbf{v}_d,\qquad I = \int \mathbf{J}\cdot d\mathbf{A} \] \(n\) = number of free carriers per cubic metre (\(8.49\times10^{28}\ \text{m}^{-3}\) for copper), \(q\) = charge per carrier, \(v_d\) = drift speed, \(A\) = cross-sectional area. The current density \(\mathbf{J}\) (A/m²) is current per unit area — a vector at each point. For electrons \(q = -e\) and \(\mathbf{v}_d\) points against \(\mathbf{E}\), so \(\mathbf{J} = nq\mathbf{v}_d\) still points along \(\mathbf{E}\): two minus signs cancel.

How slow? A 12-gauge copper house wire (diameter 2.05 mm, so \(A = 3.31\times10^{-6}\ \text{m}^2\)) carrying 10 A:

\[ v_d = \frac{I}{neA} = \frac{10}{(8.49\times10^{28})(1.60\times10^{-19})(3.31\times10^{-6})} = 2.2\times10^{-4}\ \text{m/s} \] About a fifth of a millimetre per second — slower than a snail.

Common mistake

“Then why does the light come on instantly?” Because the field — the signal — sets up along the whole wire at nearly the speed of light, and every electron in the wire starts drifting at once. The electrons near the bulb are already in the bulb. Think of a hose already full of water: open the tap and water comes out the far end at once, even though no single drop travelled the hose. Do not confuse drift speed, random speed and signal speed: they differ by about twelve orders of magnitude.

EE connection

Current density, not current, is what limits a conductor. Copper PCB traces and bond wires are rated in A/mm²; too much \(J\) heats the trace and, at very high \(J\), the electron “wind” physically pushes metal atoms along (electromigration), which eventually opens on-chip interconnects. Chip designers check \(J\) on every wire of a layout.

2 · Resistivity and Ohm's law

Collisions limit the drift: the harder you push (larger \(E\)), the faster the drift, in proportion. That proportionality is Ohm's law at the level of a single point in the material:

\[ \mathbf{J} = \sigma\mathbf{E}\qquad\Longleftrightarrow\qquad \mathbf{E} = \rho\mathbf{J},\qquad \rho = \frac{1}{\sigma} \] Microscopic Ohm's law. \(\sigma\) = conductivity (S/m), \(\rho\) = resistivity (\(\Omega\cdot\text{m}\)). Both are properties of the material, not of the shape. Copper: \(\rho = 1.68\times10^{-8}\ \Omega\cdot\text{m}\); silicon (pure): about \(2\times10^{3}\ \Omega\cdot\text{m}\); glass: \(10^{10}\)–\(10^{14}\). A range of more than 20 orders of magnitude — the widest of any physical property.

A simple model (Drude) makes this concrete: an electron accelerates at \(eE/m_e\) for an average time \(\tau\) between collisions, so \(v_d = eE\tau/m_e\), which gives \(\sigma = ne^2\tau/m_e\). For copper that requires \(\tau\approx 2.5\times10^{-14}\) s — about 25 femtoseconds between collisions.

Now take a uniform wire of length \(L\) and area \(A\). The field is \(E = V/L\) (Unit 3: potential difference = field × distance) and \(J = I/A\). Substitute into \(E = \rho J\):

\[ \frac{V}{L} = \rho\frac{I}{A}\quad\Longrightarrow\quad V = IR,\qquad R = \frac{\rho L}{A} \] Macroscopic Ohm's law and resistance in ohms (\(\Omega\) = V/A). Longer wire, more resistance (resistors in series); fatter wire, less resistance (paths in parallel). 10 m of that 12-gauge wire: \(R = (1.68\times10^{-8})(10)/(3.31\times10^{-6}) = 0.051\ \Omega\).

Common mistake

\(V = IR\) is not a law of nature the way Coulomb's law is. It is a property of ohmic materials: \(R\) is constant, so the \(I\)–\(V\) graph is a straight line through the origin. A diode, an LED, or a light bulb filament (whose \(R\) rises as it heats) is non-ohmic. The definition \(R = V/I\) still works at any single operating point, but \(R\) changes with \(V\).

Temperature dependence

\[ \rho(T) \approx \rho_0\left[1 + \alpha\,(T - T_0)\right] \] \(\rho_0\) is the resistivity at reference temperature \(T_0\) (usually 20 °C) and \(\alpha\) the temperature coefficient of resistivity (copper: \(3.93\times10^{-3}\ /^\circ\text{C}\)). Metals: hotter lattice → more collisions → \(\alpha > 0\). Semiconductors: hotter → many more free carriers → \(\rho\) falls, \(\alpha < 0\). The same formula applies to \(R\) when the geometry is fixed.

EE connection

A copper motor winding at 80 °C has \(R = R_0[1 + 0.00393(60)] = 1.24R_0\) — 24 % more than at 20 °C. Engineers measure winding temperature exactly this way (resistance rise method). Platinum RTDs (Pt100: 100 Ω at 0 °C) and NTC thermistors in battery packs use the same physics as temperature sensors.

3 · Electrical power

A charge \(dQ\) falling through a potential difference \(V\) loses potential energy \(dQ\,V\) (Unit 3). Divide by \(dt\):

\[ P = IV = I^2R = \frac{V^2}{R} \] \(P = IV\) holds for any element — motor, battery, LED: it is the rate at which energy enters or leaves the circuit at that element. The forms \(I^2R\) and \(V^2/R\) are only for resistors (they use \(V = IR\)). In a resistor the energy becomes heat — Joule heating.

Which formula?

Pick the one whose quantity is fixed in the situation. Two resistors in series share the same \(I\), so compare \(I^2R\): the bigger resistor gets more power. In parallel they share \(V\), so compare \(V^2/R\): the smaller resistor gets more power. Both are true; they answer different questions.

EE connection

Why do power lines run at hundreds of kilovolts? To deliver power \(P = IV\) with a small current, because the line loss is \(I^2R_{\text{line}}\). Doubling the voltage halves \(I\) and quarters the loss. Unit 9 shows how transformers make that possible. And every resistor on a board has a power rating (⅛ W, ¼ W …): exceed \(I^2R\) and it cooks.

4 · EMF and internal resistance

Inside a resistor, charges lose potential energy. Something must lift them back up, otherwise the current would die. That something is a source of EMF (battery, generator, solar cell, power supply). Inside a battery, chemistry pushes positive charge from the − terminal to the + terminal, against the electric field — like a pump lifting water uphill.

\[ \mathcal{E} = \frac{dW_{\text{non-electric}}}{dQ}\qquad [\text{V}] \] The EMF \(\mathcal{E}\) is the work done per coulomb by the non-electrostatic force in the source. Despite the name (“electromotive force”) it is a voltage, not a force.

A real source = an ideal EMF \(\mathcal{E}\) in series with an internal resistance \(r\). You can only touch terminals \(a\) and \(b\), so you only ever see \(V_{ab} = \mathcal{E} - Ir\). The long plate of the battery symbol is the + terminal.

\[ V_{ab} = \mathcal{E} - Ir,\qquad I = \frac{\mathcal{E}}{R + r} \] \(V_{ab}\) = terminal voltage; \(r\) = internal resistance. With no current (open circuit) a voltmeter reads \(\mathcal{E}\). The more current you draw, the more the terminal voltage sags. A short circuit (\(R = 0\)) gives \(I = \mathcal{E}/r\), all dissipated inside the battery.

Maximum power transfer

\[ P_R = I^2R = \frac{\mathcal{E}^2R}{(R+r)^2},\qquad \frac{dP_R}{dR} = \mathcal{E}^2\frac{(R+r) - 2R}{(R+r)^3} = 0 \;\Rightarrow\; R = r,\qquad P_{\max} = \frac{\mathcal{E}^2}{4r} \] The load receives the most power when its resistance matches the source's internal resistance. At that point the efficiency is only 50 % — half the power heats the source.

EE connection

Every real source — a battery, a sensor, an amplifier output, an antenna — can be modelled as an ideal source plus an internal resistance (the Thévenin equivalent, coming in your circuits course). Signal engineers match impedances (50 Ω RF cables, 50 Ω instruments) to get the most power into the load; power engineers do the opposite — they keep \(R_{\text{load}}\gg r\) because they care about efficiency, not maximum power.

5 · Resistors in series and parallel

\[ \text{Series: } R_{\text{eq}} = R_1 + R_2 + \cdots\qquad\qquad \text{Parallel: } \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots \] Series = one path, same current through each, voltages add. Parallel = both ends shared, same voltage across each, currents add. For two in parallel, \(R_{\text{eq}} = R_1R_2/(R_1 + R_2)\) (“product over sum”). Note the pattern is the opposite of capacitors (Unit 4).

Why: in series, \(V = IR_1 + IR_2 = I(R_1 + R_2)\). In parallel, \(I = V/R_1 + V/R_2 = V(1/R_1 + 1/R_2)\). Two sanity checks that catch most errors: a series combination is larger than its largest member; a parallel combination is smaller than its smallest member.

\[ \text{Voltage divider: } V_2 = V\frac{R_2}{R_1 + R_2}\qquad\qquad \text{Current divider: } I_2 = I\frac{R_1}{R_1 + R_2} \] In series the bigger resistor takes the bigger share of voltage. In parallel the current divides inversely — the other resistor appears on top, so the smaller resistor takes the bigger current.

Reducing a network

Redraw it. Two resistors are in series only if nothing else connects at the node between them; they are in parallel only if both of their ends are on the same two nodes. Collapse the innermost pair, redraw, repeat. Example: \(6\ \Omega\parallel 3\ \Omega = 2\ \Omega\); plus \(4\ \Omega\) in series \(= 6\ \Omega\); that in parallel with \(12\ \Omega\) gives \(4\ \Omega\). Some networks (a bridge, for instance) never reduce this way — that is what Kirchhoff's rules are for.

6 · Kirchhoff's rules

Two conservation laws, written for circuits, solve every DC circuit:

\[ \text{Junction rule (KCL):}\quad \sum I_{\text{in}} = \sum I_{\text{out}}\qquad\qquad \text{Loop rule (KVL):}\quad \sum_{\text{around a loop}} \Delta V = 0 \] KCL is conservation of charge: charge does not pile up at a junction. KVL is conservation of energy (equivalently, the electrostatic field is conservative, \(\oint\mathbf{E}\cdot d\mathbf{l} = 0\)): walk around any closed loop and you return to the same potential, so all the rises and drops sum to zero.

Element	Walking direction	\(\Delta V\)
Resistor \(R\)	Same direction as your assumed current \(I\)	\(-IR\) (downhill)
Resistor \(R\)	Against your assumed current	\(+IR\)
EMF \(\mathcal{E}\)	From − terminal to + terminal	\(+\mathcal{E}\)
EMF \(\mathcal{E}\)	From + terminal to − terminal	\(-\mathcal{E}\)

The recipe

Label a current in every branch (a branch = a path between two junctions) with an arrow. Guess the direction — a wrong guess just gives a negative answer.
KCL at junctions. With \(j\) junctions you get \(j - 1\) independent equations.
KVL around enough loops to have as many equations as unknown currents. Each new loop must include at least one branch not yet used.
Solve, then check: substitute into a loop you did not use, and check that power delivered equals power dissipated.

A 14 V charger (internal resistance 1 Ω) connected to a 7 V battery (internal resistance 2 Ω), with a 4 Ω load across them. Assumed directions: \(I_1\) up the left, \(I_2\) up the right, \(I_3\) down the middle. The long plate is + in both sources.

Solve it. Unknowns \(I_1, I_2, I_3\). KCL at junction \(a\):

\[ I_1 + I_2 = I_3 \]

KVL, loop 1, walking clockwise from \(b\): up through the 14 V source (− to +, \(+14\)), up through \(R_1\) with \(I_1\) (\(-1\cdot I_1\)), down through \(R_3\) with \(I_3\) (\(-4I_3\)). Loop 2, walking counterclockwise from \(b\): up through the 7 V battery (\(+7\)), up through \(R_2\) with \(I_2\) (\(-2I_2\)), down through \(R_3\) (\(-4I_3\)):

\[ 14 - I_1 - 4I_3 = 0,\qquad 7 - 2I_2 - 4I_3 = 0 \] \[ I_1 = 14 - 4I_3,\quad I_2 = \tfrac{7}{2} - 2I_3 \;\Rightarrow\; 14 - 4I_3 + \tfrac72 - 2I_3 = I_3 \;\Rightarrow\; 7I_3 = 17.5 \] \[ I_3 = 2.5\ \text{A},\qquad I_1 = 4.0\ \text{A},\qquad I_2 = -1.5\ \text{A} \] \(I_2\) is negative: 1.5 A actually flows down through the 7 V battery, into its + terminal. The battery is being charged. That is the whole point of the circuit — and the sign told us, without any extra thinking.

Check in every loop (including the outer loop, which we never used):

\[ \text{Loop 1: } 14 - 4.0 - 4(2.5) = 0\ \checkmark\qquad \text{Loop 2: } 7 - 2(-1.5) - 4(2.5) = 7 + 3 - 10 = 0\ \checkmark \] \[ \text{Outer (up left, down right): } 14 - 1(4.0) + 2(-1.5) - 7 = 14 - 4 - 3 - 7 = 0\ \checkmark \] \[ \text{Power: } \underbrace{14(4.0)}_{56\ \text{W in}} = \underbrace{7(1.5)}_{10.5\ \text{charging}} + \underbrace{4.0^2(1)}_{16} + \underbrace{1.5^2(2)}_{4.5} + \underbrace{2.5^2(4)}_{25} = 56\ \text{W}\ \checkmark \]

Common mistakes

Sign of \(IR\) depends on your walking direction relative to the assumed current — not on which way current “really” flows.
Sign of \(\mathcal{E}\) depends only on which terminal you meet first — not on the current direction.
Flipping a negative answer mid-solution. Keep the negative number all the way through; interpret it at the end.
Writing more loop equations than you need and getting a dependent (useless) set — each loop must add a new branch.

EE connection

This is exactly what a car's alternator does to its 12 V battery, and what a laptop's charger does while the laptop runs. In your circuits course you will systematise KVL as mesh analysis and KCL as nodal analysis; circuit simulators like SPICE solve the nodal equations for thousands of nodes as a matrix. Worked example 3 below uses mesh currents.

7 · Ammeters and voltmeters

A meter can only measure what is happening if it does not change it much. That single idea decides how each meter is built and connected.

Meter	Connect it	Ideal resistance	Why
Ammeter	In series (break the circuit, let current go through it)	\(R_A\to 0\)	Adds \(R_A\) to the branch; must not reduce the current.
Voltmeter	In parallel (across the two points)	\(R_V\to\infty\)	Draws \(V/R_V\); must not steal current from the element.

A classic analog meter starts from a galvanometer — a coil in a magnet (Unit 6!) that deflects fully at a small current \(I_g\) and has resistance \(R_g\). To make an ammeter, put a tiny shunt \(R_s\) in parallel so most of the current bypasses the coil: \(R_s = I_gR_g/(I - I_g)\). To make a voltmeter, put a big resistor in series with the coil.

Common mistake — that blows fuses

Connecting an ammeter across a battery (in parallel) shorts the battery through a near-zero resistance. Multimeters have a fuse in the current jack for exactly this reason. Before measuring voltage, check the red lead is in the V jack, not the A jack.

EE connection: loading

A typical DMM has \(R_V = 10\ \text{M}\Omega\). Measure the midpoint of a divider of two \(1\ \text{M}\Omega\) resistors across 10 V: the meter in parallel with the lower resistor makes \(0.909\ \text{M}\Omega\), so you read \(10\times 0.909/1.909 = 4.76\) V instead of 5.00 V — a 5 % error caused by the meter. Oscilloscope probes (\(1\ \text{M}\Omega\) or \(10\ \text{M}\Omega\) with a 10× probe) load circuits the same way. Rule of thumb: the meter resistance should be at least 100× the resistance it sits across.

8 · RC circuits

Put a capacitor in a circuit and the currents stop being constant. A capacitor cannot change its charge instantly — charge must flow in through the resistor, and the more charge is already there, the more it pushes back. Result: exponential approach. Two limits to remember first:

Just after switching (\(t = 0^+\)), an uncharged capacitor has \(V_C = q/C = 0\): it behaves like a wire.
After a long time (\(t\to\infty\)), the current into it is zero: it behaves like a break (open circuit).

Charging: set up and solve the differential equation

A switch connects an EMF \(\mathcal{E}\), a resistor \(R\) and an uncharged capacitor \(C\) in series at \(t = 0\). Let \(q(t)\) be the capacitor charge and \(i = dq/dt\) the current. KVL around the loop:

\[ \mathcal{E} - iR - \frac{q}{C} = 0\quad\Longrightarrow\quad R\frac{dq}{dt} = \mathcal{E} - \frac{q}{C} = \frac{C\mathcal{E} - q}{C} \] A first-order linear ODE. Separate the variables: all \(q\) on the left, all \(t\) on the right.

\[ \int_0^{q}\frac{dq'}{C\mathcal{E} - q'} = \int_0^{t}\frac{dt'}{RC}\quad\Longrightarrow\quad -\ln\frac{C\mathcal{E} - q}{C\mathcal{E}} = \frac{t}{RC} \] \[ q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right),\qquad i(t) = \frac{dq}{dt} = \frac{\mathcal{E}}{R}e^{-t/RC},\qquad V_C(t) = \mathcal{E}\left(1 - e^{-t/RC}\right) \] Check the limits: at \(t = 0\), \(q = 0\) and \(i = \mathcal{E}/R\) (capacitor acts as a wire); as \(t\to\infty\), \(q\to C\mathcal{E}\) and \(i\to 0\) (open circuit). ✔ fits both pictures above.

\[ \tau = RC\qquad [\Omega\cdot\text{F} = \text{s}] \] The time constant. After one \(\tau\) the capacitor is \(1 - e^{-1} = 63.2\%\) charged; after \(5\tau\), \(99.3\%\) — engineers call that “fully charged”. \(10\ \text{k}\Omega\times 100\ \mu\text{F} = 1.0\) s.

Charging: the capacitor voltage \(V_C/\mathcal{E} = 1 - e^{-t/\tau}\) (accent) and the current \(i/(\mathcal{E}/R) = e^{-t/\tau}\) (blue). At \(t = \tau\) the voltage has covered 63 % of the way and the current has fallen to 37 %. The two always add to 100 % because \(V_R + V_C = \mathcal{E}\).

Discharging

Remove the battery and connect a charged capacitor (\(q = Q_0\)) across \(R\). Now KVL is \(-iR - q/C = 0\) with \(i = dq/dt\):

\[ \frac{dq}{dt} = -\frac{q}{RC}\quad\Longrightarrow\quad q(t) = Q_0e^{-t/RC},\qquad i(t) = -\frac{Q_0}{RC}e^{-t/RC},\qquad V_C = V_0e^{-t/RC} \] The minus sign in \(i\) only says the current now flows out of the positive plate — opposite to the charging direction. Time to fall to a fraction \(f\): \(t = \tau\ln(1/f)\); to 10 %, \(t = \tau\ln 10 = 2.30\tau\); to half, \(t = \tau\ln 2 = 0.693\tau\).

Energy

\[ W_{\text{battery}} = \mathcal{E}Q = C\mathcal{E}^2,\qquad U_C = \tfrac12C\mathcal{E}^2,\qquad W_R = \int_0^\infty i^2R\,dt = \frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2} = \tfrac12C\mathcal{E}^2 \] Charging through a resistor always wastes exactly half the energy the battery supplies, whatever \(R\) is. A small \(R\) gives a big current for a short time; a big \(R\) a small current for a long time — the heat is the same. (The integral: \(\int_0^\infty e^{-2t/\tau}dt = \tau/2\).)

Common mistake

\(\tau = RC\) uses the resistance the capacitor sees. If several resistors surround it, turn off the sources (replace an ideal battery by a wire), look back into the circuit from the capacitor's terminals, and find the equivalent resistance. Using the “total” circuit resistance instead is the most common RC error.

EE connection

Every wire on a chip has resistance and every gate input has capacitance, so every digital edge is an RC charging curve: the 10 %–90 % rise time is \(t_{90} - t_{10} = \tau\ln 10 - \tau\ln\tfrac{10}{9} = \tau\ln 9\approx 2.2\tau\). RC is also the timing element in the 555 timer (it charges between \(\tfrac13\) and \(\tfrac23\) of the supply, which takes \(\tau\ln2\)), the simplest low-pass filter (cutoff \(f_c = 1/(2\pi RC)\), Unit 9), switch debouncers and power-on reset circuits.

Interactive lab RC circuit on an oscilloscope Open it in Interactive Laboratories →

9 · Worked examples

1. Voltage drop in an extension cord

A 15 m extension cord has two 14-gauge copper conductors (diameter 1.628 mm). It feeds a 12 A space heater. Find the resistance of the cord, the voltage lost in it and the power wasted.

\[ A = \pi r^2 = \pi(0.814\times10^{-3}\ \text{m})^2 = 2.08\times10^{-6}\ \text{m}^2 \] \[ R_{\text{one wire}} = \frac{\rho L}{A} = \frac{(1.68\times10^{-8})(15)}{2.08\times10^{-6}} = 0.121\ \Omega,\qquad R_{\text{cord}} = 2(0.121) = 0.242\ \Omega \]

Current goes out on one conductor and back on the other, so the path length is 30 m — the factor of 2 is the usual trap.

\[ \Delta V = IR = 12(0.242) = \mathbf{2.9\ V},\qquad P = I^2R = 144(0.242) = \mathbf{35\ W} \]

35 W of heat spread along the cord — fine. But coil the cord tightly on its reel and that heat cannot escape, which is why reels warn you to unwind them fully under load.

2. Measuring a battery's internal resistance

A car battery reads 12.6 V with nothing connected and 11.8 V while delivering 20 A. Find \(\mathcal{E}\), \(r\), the load resistance, and the most power the battery could deliver to any load.

Open circuit: \(I = 0\), so \(V_{ab} = \mathcal{E} = 12.6\) V. Under load:

\[ r = \frac{\mathcal{E} - V_{ab}}{I} = \frac{12.6 - 11.8}{20} = \mathbf{0.040\ \Omega},\qquad R = \frac{V_{ab}}{I} = \frac{11.8}{20} = \mathbf{0.59\ \Omega} \] \[ P_{\max} = \frac{\mathcal{E}^2}{4r} = \frac{12.6^2}{4(0.040)} = \frac{158.76}{0.16} \approx \mathbf{990\ W}\quad(\text{at } R = r = 0.040\ \Omega) \]

Check: \(\mathcal{E}/(R + r) = 12.6/0.63 = 20\) A ✔. Battery testers do exactly this two-point measurement: a battery whose \(r\) has grown with age still shows 12.6 V open-circuit but sags badly under a starter-motor load.

3. Two loops with mesh currents

Same layout as the figure in Section 6, new values: left branch \(\mathcal{E}_1 = 9\) V (+ at top) in series with \(R_1 = 3\ \Omega\); middle branch \(R_3 = 6\ \Omega\); right branch \(R_2 = 2\ \Omega\) in series with \(\mathcal{E}_2 = 3\) V whose + terminal is at the bottom. Find every branch current.

Mesh currents: let \(i_L\) circulate clockwise in the left window and \(i_R\) clockwise in the right window. Then the left branch carries \(i_L\) upward, the right branch carries \(i_R\) downward, and the shared middle resistor carries \(i_L - i_R\) downward. KCL is automatically satisfied — that is the advantage of meshes.

Left mesh, clockwise: up through \(\mathcal{E}_1\) (\(+9\)), up through \(R_1\) (\(-3i_L\)), down through \(R_3\) with net current \(i_L - i_R\) (\(-6(i_L - i_R)\)). Right mesh, clockwise: up through \(R_3\), where the net clockwise current is \(i_R - i_L\) (\(-6(i_R - i_L)\)), down through \(R_2\) (\(-2i_R\)), down through \(\mathcal{E}_2\) from − to + (\(+3\)):

\[ 9 - 9i_L + 6i_R = 0,\qquad 6i_L - 8i_R + 3 = 0 \] \[ i_L = 1 + \tfrac23 i_R\;\Rightarrow\; 6 + 4i_R - 8i_R + 3 = 0\;\Rightarrow\; i_R = 2.25\ \text{A},\quad i_L = 2.50\ \text{A} \]

Branch currents: left \(\mathbf{2.50\ A}\) up; right \(\mathbf{2.25\ A}\) down; middle \(i_L - i_R = \mathbf{0.25\ A}\) down.

Substitute back into every loop:

\[ \text{Left: } 9 - 3(2.50) - 6(0.25) = 9 - 7.5 - 1.5 = 0\ \checkmark \qquad \text{Right: } -6(-0.25) - 2(2.25) + 3 = 1.5 - 4.5 + 3 = 0\ \checkmark \] \[ \text{Outer, clockwise: } 9 - 3(2.50) - 2(2.25) + 3 = 9 - 7.5 - 4.5 + 3 = 0\ \checkmark \] \[ \text{Power: } 9(2.50) + 3(2.25) = 29.25\ \text{W} = 3(2.5^2) + 6(0.25^2) + 2(2.25^2) = 18.75 + 0.375 + 10.125\ \checkmark \]

Both sources push current clockwise around the outer loop, so both deliver power; very little current is left for the middle resistor.

4. A voltmeter that changes what it measures

Two \(1.00\ \text{M}\Omega\) resistors in series across 10.0 V. A voltmeter with \(R_V = 10.0\ \text{M}\Omega\) is placed across the lower one. What does it read, and what is the percentage error?

\[ R_\parallel = \frac{(1.00)(10.0)}{1.00 + 10.0}\ \text{M}\Omega = 0.909\ \text{M}\Omega,\qquad V_{\text{read}} = 10.0\cdot\frac{0.909}{1.00 + 0.909} = \mathbf{4.76\ V} \]

True value (no meter) 5.00 V, so the error is \(-4.8\%\). With \(1\ \text{k}\Omega\) resistors instead, the same meter would give \(R_\parallel = 999.9\ \Omega\) and an error of 0.005 % — loading only matters when the circuit's resistances approach the meter's.

5. Charging an RC circuit

\(\mathcal{E} = 9.0\) V, \(R = 10\ \text{k}\Omega\), \(C = 100\ \mu\text{F}\), capacitor initially uncharged. Find \(\tau\), the initial current, \(V_C\) at \(t = 2.0\) s, and the energy bookkeeping once fully charged.

\[ \tau = RC = (1.0\times10^4)(1.0\times10^{-4}) = \mathbf{1.0\ s},\qquad i(0) = \frac{\mathcal{E}}{R} = \frac{9.0}{1.0\times10^4} = \mathbf{0.90\ mA} \] \[ V_C(2.0\ \text{s}) = 9.0\left(1 - e^{-2.0}\right) = 9.0(0.8647) = \mathbf{7.8\ V} \] \[ W_{\text{battery}} = C\mathcal{E}^2 = (1.0\times10^{-4})(81) = 8.1\ \text{mJ},\quad U_C = \tfrac12C\mathcal{E}^2 = \mathbf{4.05\ mJ},\quad W_R = \mathbf{4.05\ mJ} \]

Half stored, half turned to heat in \(R\), as promised. At 2τ the capacitor is 86 % charged; at 5τ = 5 s it is 99.3 %.

6. Bleeder resistor: a safety discharge

A power supply's \(1000\ \mu\text{F}\) filter capacitor charges to 170 V (the peak of 120 V AC mains). A \(100\ \text{k}\Omega\) “bleeder” resistor is permanently across it. After unplugging, how long until \(V_C\) falls to 50 V? How much energy was stored?

\[ \tau = RC = (1.0\times10^5)(1.0\times10^{-3}) = 100\ \text{s} \] \[ 50 = 170\,e^{-t/\tau}\;\Rightarrow\; t = \tau\ln\frac{170}{50} = 100\ln 3.4 = 100(1.224) = \mathbf{122\ s} \] \[ U_0 = \tfrac12CV_0^2 = \tfrac12(1.0\times10^{-3})(170)^2 = \mathbf{14.5\ J} \]

Two minutes. That is why service manuals say “wait several minutes after unplugging” — and why you should never trust it: if the bleeder has failed open, \(\tau\) is effectively hours. Measure before touching. (Power drawn by the bleeder in normal operation: \(170^2/10^5\approx 0.29\) W — the price of safety.)

Checklist

I can relate \(I\), \(J\), \(n\), \(q\), \(v_d\) and explain why drift is slow but signals are fast.
I know \(\mathbf{J} = \sigma\mathbf{E}\) is the point form of \(V = IR\), and \(R = \rho L/A\).
I can use \(\rho = \rho_0[1 + \alpha(T - T_0)]\) and know the sign of \(\alpha\) for metals vs semiconductors.
I use \(P = IV\) for anything and \(I^2R\), \(V^2/R\) only for resistors.
I model a real source as \(\mathcal{E}\) plus \(r\), and know maximum power transfer happens at \(R = r\).
I can reduce series/parallel networks and use voltage and current dividers.
I can write KCL and KVL with consistent signs, solve, and check by substituting into every loop and balancing power.
I know ammeters go in series (low \(R\)), voltmeters in parallel (high \(R\)), and can estimate loading error.
I can derive \(q(t)\) for charging and discharging from the ODE, use \(\tau = RC\), and show half the energy is lost while charging.