Practice Exam 5

Current, Resistance & DC Circuits

Questions10

Time90 min

Points100

CalculatorAllowed

Use \(e = 1.60\times10^{-19}\) C, copper \(n = 8.49\times10^{28}\ \text{m}^{-3}\), \(\rho_{\text{Cu}} = 1.68\times10^{-8}\ \Omega\cdot\text{m}\), \(\alpha_{\text{Cu}} = 3.93\times10^{-3}\ /^\circ\text{C}\). Batteries are ideal unless an internal resistance is given.

1Current density & drift10 pts

A copper wire of diameter 1.00 mm carries 2.00 A. (a) Find the current density \(J\) and the field \(E\) inside the wire. (b) Find the drift speed and how long an electron takes to drift 1.00 m. (c) The same current flows through a section of half the diameter. By what factor do \(J\) and \(v_d\) change? (d) Why does a lamp still turn on “instantly”?

Solution

\[ A = \pi(0.500\times10^{-3})^2 = 7.85\times10^{-7}\ \text{m}^2,\qquad J = \frac{I}{A} = \frac{2.00}{7.85\times10^{-7}} = \mathbf{2.55\times10^{6}\ A/m^2} \] \[ E = \rho J = (1.68\times10^{-8})(2.55\times10^{6}) = \mathbf{0.0428\ V/m} \] \[ v_d = \frac{J}{ne} = \frac{2.55\times10^{6}}{(8.49\times10^{28})(1.60\times10^{-19})} = \mathbf{1.87\times10^{-4}\ m/s},\qquad t = \frac{1.00}{1.87\times10^{-4}} = 5.3\times10^{3}\ \text{s}\approx \mathbf{89\ min} \]

(c) Half the diameter is a quarter of the area. Same \(I\), so \(J = I/A\) and \(v_d = J/(ne)\) both become 4 times larger — current is conserved, so carriers must speed up through the narrow part (like water in a nozzle).

(d) The field that drives the electrons is established along the whole wire at nearly the speed of light; electrons everywhere in the circuit, including those already in the filament, start drifting together.

2Resistivity & temperature8 pts

A motor winding is copper wire of diameter 0.50 mm with resistance 2.40 Ω at 20 °C. (a) How long is the wire? (b) After an hour of running, the winding measures 2.88 Ω. Estimate its temperature.

Solution

\[ A = \pi(0.25\times10^{-3})^2 = 1.96\times10^{-7}\ \text{m}^2,\qquad L = \frac{RA}{\rho} = \frac{(2.40)(1.96\times10^{-7})}{1.68\times10^{-8}} = \mathbf{28\ m} \] \[ \frac{R}{R_0} = 1 + \alpha\Delta T\;\Rightarrow\;\Delta T = \frac{2.88/2.40 - 1}{3.93\times10^{-3}} = \frac{0.200}{3.93\times10^{-3}} = 50.9\ ^\circ\text{C},\qquad T\approx \mathbf{71\ ^\circ C} \]

Thermal expansion changes \(L\) and \(A\) by only ~0.1 %, so the 20 % rise is almost entirely the resistivity. This “resistance rise” method is the standard way to measure the average temperature of a winding you cannot reach with a thermometer.

3EMF, internal resistance & power10 pts

A battery with \(\mathcal{E} = 12.0\) V and \(r = 0.50\ \Omega\) drives a \(5.5\ \Omega\) load. (a) Find the current, terminal voltage, power to the load, power lost inside, and efficiency. (b) What load resistance draws the maximum power, how much, and at what efficiency?

Solution

\[ I = \frac{\mathcal{E}}{R + r} = \frac{12.0}{6.0} = \mathbf{2.0\ A},\qquad V_{ab} = \mathcal{E} - Ir = 12.0 - 1.0 = \mathbf{11.0\ V} \] \[ P_R = I^2R = 4(5.5) = \mathbf{22\ W},\qquad P_r = I^2r = 4(0.5) = \mathbf{2.0\ W},\qquad \eta = \frac{22}{24} = \mathbf{92\%} \]

Check: \(\mathcal{E}I = 24\) W \(= 22 + 2\) ✔.

\[ R = r = 0.50\ \Omega:\qquad P_{\max} = \frac{\mathcal{E}^2}{4r} = \frac{144}{2.0} = \mathbf{72\ W},\qquad \eta = \mathbf{50\%} \]

Maximum power and maximum efficiency are different goals: the matched load gets three times more power but wastes as much as it uses.

4Series–parallel reduction10 pts

A 10.0 V battery drives a \(1\ \Omega\) resistor in series with a parallel block. The block has two branches: branch A is a single \(12\ \Omega\) resistor; branch B is a \(4\ \Omega\) resistor in series with a \(6\ \Omega\parallel 3\ \Omega\) pair. Find the equivalent resistance, the battery current and the current through the \(3\ \Omega\) resistor.

Solution

\[ 6\parallel3 = \frac{18}{9} = 2\ \Omega,\quad \text{branch B} = 4 + 2 = 6\ \Omega,\quad 12\parallel 6 = \frac{72}{18} = 4\ \Omega,\quad R_{\text{eq}} = 1 + 4 = \mathbf{5\ \Omega} \] \[ I = \frac{10.0}{5} = \mathbf{2.0\ A},\qquad V_{\text{block}} = 2.0(4) = 8.0\ \text{V} \] \[ I_B = \frac{8.0}{6} = \tfrac43\ \text{A},\qquad V_{6\parallel3} = \tfrac43(2) = \tfrac83\ \text{V},\qquad I_{3\Omega} = \frac{8/3}{3} = \tfrac89\ \text{A} = \mathbf{0.89\ A} \]

Checks: \(I_A = 8/12 = \tfrac23\) A and \(\tfrac23 + \tfrac43 = 2\) A ✔. Power: \(1(4) + 12(\tfrac49) + 4(\tfrac{16}{9}) + 3(\tfrac{64}{81}) + 6(\tfrac{16}{81}) = 4 + 5.33 + 7.11 + 2.37 + 1.19 = 20.0\) W \(= 10(2.0)\) ✔.

5Kirchhoff's rules (two loops)14 pts

Two junctions \(a\) (top) and \(b\) (bottom) are joined by three branches. Left: \(\mathcal{E}_1 = 18\) V (+ terminal toward \(a\)) in series with \(R_1 = 6\ \Omega\). Middle: \(R_3 = 3\ \Omega\). Right: \(\mathcal{E}_2 = 12\) V (+ toward \(a\)) in series with \(R_2 = 2\ \Omega\). Find all three currents and \(V_a - V_b\), verify by substituting into every loop, and check the power balance.

Solution

Assume \(I_1\) up the left, \(I_2\) up the right, \(I_3\) down the middle. KCL at \(a\): \(I_1 + I_2 = I_3\). KVL (left window, then right window, each walked up the outer branch and down the middle):

\[ 18 - 6I_1 - 3I_3 = 0,\qquad 12 - 2I_2 - 3I_3 = 0 \] \[ I_1 = 3 - \tfrac12I_3,\quad I_2 = 6 - \tfrac32I_3\;\Rightarrow\; 9 - 2I_3 = I_3\;\Rightarrow\; I_3 = \mathbf{3.0\ A},\quad I_1 = \mathbf{1.5\ A},\quad I_2 = \mathbf{1.5\ A} \] \[ V_a - V_b = I_3R_3 = \mathbf{9.0\ V} \]

All positive, so the assumed directions were right: both sources discharge into the middle resistor.

\[ \text{Left: } 18 - 6(1.5) - 3(3.0) = 18 - 9 - 9 = 0\ \checkmark\qquad \text{Right: } 12 - 2(1.5) - 3(3.0) = 12 - 3 - 9 = 0\ \checkmark \] \[ \text{Outer (up left, down right): } 18 - 6(1.5) + 2(1.5) - 12 = 18 - 9 + 3 - 12 = 0\ \checkmark \] \[ \text{Power: } 18(1.5) + 12(1.5) = 45\ \text{W} = 6(1.5^2) + 2(1.5^2) + 3(3.0^2) = 13.5 + 4.5 + 27\ \checkmark \]

Also check \(V_{ab}\) along each outer branch: left \(18 - 6(1.5) = 9\) V ✔, right \(12 - 2(1.5) = 9\) V ✔.

6Meters & loading · EE application10 pts

(a) Two \(100\ \text{k}\Omega\) resistors in series sit across 20.0 V. A meter with \(R_V = 2.0\ \text{M}\Omega\) measures the voltage across one of them. What does it read, and what is the error? (b) A galvanometer reads full scale at 1.00 mA and has resistance \(50.0\ \Omega\). What shunt turns it into a 0-to-1.00 A ammeter, and is it connected in series or parallel with the coil?

Solution

\[ R_\parallel = \frac{(1.0\times10^5)(2.0\times10^6)}{2.1\times10^6} = 95.2\ \text{k}\Omega,\qquad V = 20.0\cdot\frac{95.2}{100 + 95.2} = \mathbf{9.76\ V} \]

True value 10.00 V, error \(\mathbf{-2.4\%}\). The meter's resistance is only 20× the resistance it sits across — too close.

(b) The shunt goes in parallel with the coil. At full scale 1.00 mA flows through the coil, the other 0.999 A through the shunt, and both see the same voltage:

\[ I_gR_g = (I - I_g)R_s\;\Rightarrow\; R_s = \frac{(1.00\times10^{-3})(50.0)}{0.999} = \mathbf{0.0501\ \Omega} \]

The finished ammeter has resistance \(50.0\parallel0.0501\approx0.050\ \Omega\) — small, as an ammeter must be. It then goes in series with the circuit.

7RC charging: derivation12 pts

(a) Starting from Kirchhoff's loop rule, derive \(V_C(t)\) for an uncharged capacitor \(C\) connected at \(t = 0\) to an EMF \(\mathcal{E}\) through a resistor \(R\). (b) For \(\mathcal{E} = 5.0\) V, \(R = 2.0\ \text{k}\Omega\), \(C = 1.0\ \mu\text{F}\), find \(\tau\), the initial current, and the time for \(V_C\) to reach 4.0 V.

Solution

\[ \mathcal{E} - R\frac{dq}{dt} - \frac{q}{C} = 0\;\Rightarrow\;\frac{dq}{C\mathcal{E} - q} = \frac{dt}{RC}\;\Rightarrow\;-\ln\!\left(\frac{C\mathcal{E} - q}{C\mathcal{E}}\right) = \frac{t}{RC} \] \[ q = C\mathcal{E}\left(1 - e^{-t/RC}\right),\qquad V_C = \frac{q}{C} = \mathcal{E}\left(1 - e^{-t/RC}\right) \]

The lower limits used \(q(0) = 0\). Checks: \(V_C(0) = 0\), \(V_C(\infty) = \mathcal{E}\) ✔.

\[ \tau = RC = (2.0\times10^3)(1.0\times10^{-6}) = \mathbf{2.0\ ms},\qquad i(0) = \frac{\mathcal{E}}{R} = \mathbf{2.5\ mA} \] \[ 4.0 = 5.0\left(1 - e^{-t/\tau}\right)\;\Rightarrow\; e^{-t/\tau} = 0.20\;\Rightarrow\; t = \tau\ln 5 = 2.0(1.609) = \mathbf{3.2\ ms} \]

8RC timing · EE application8 pts

A timing circuit uses \(R = 47\ \text{k}\Omega\) and \(C = 10\ \mu\text{F}\). (a) Find \(\tau\). (b) Charged to 9.0 V and then discharged through \(R\), how long until \(V_C = 3.0\) V? (c) In a 555 timer, \(C\) charges through \(R\) toward \(V_{cc}\) starting at \(\tfrac13V_{cc}\), and the output flips at \(\tfrac23V_{cc}\). Show this takes \(\tau\ln 2\) and evaluate it.

Solution

\[ \tau = (4.7\times10^4)(1.0\times10^{-5}) = \mathbf{0.47\ s},\qquad 3.0 = 9.0e^{-t/\tau}\Rightarrow t = \tau\ln3 = 0.47(1.099) = \mathbf{0.52\ s} \]

(c) Charging toward \(V_{cc}\) from \(V_0\): \(V_C = V_{cc} - (V_{cc} - V_0)e^{-t/\tau}\). With \(V_0 = \tfrac13V_{cc}\), set \(V_C = \tfrac23V_{cc}\):

\[ \tfrac13V_{cc} = \tfrac23V_{cc}\,e^{-t/\tau}\;\Rightarrow\; e^{-t/\tau} = \tfrac12\;\Rightarrow\; t = \tau\ln2 = 0.47(0.693) = \mathbf{0.33\ s} \]

Note \(V_{cc}\) cancels: the 555's timing does not depend on the supply voltage. That is the reason for the \(\tfrac13\)/\(\tfrac23\) thresholds.

9Energy in charging8 pts

A \(470\ \mu\text{F}\) capacitor is charged from 0 to 25 V by a 25 V supply through a resistor \(R\). (a) Show by integrating \(i^2R\) that the heat in \(R\) equals the energy finally stored, independent of \(R\), and give the numbers. (b) If it is instead charged in two steps — first from a 12.5 V supply, then from the 25 V supply — how much heat is produced?

Solution

\[ W_R = \int_0^\infty\left(\frac{\mathcal{E}}{R}e^{-t/RC}\right)^2R\,dt = \frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2} = \tfrac12C\mathcal{E}^2 \] \[ U_C = \tfrac12(470\times10^{-6})(25)^2 = \mathbf{0.147\ J} = W_R,\qquad W_{\text{supply}} = C\mathcal{E}^2 = \mathbf{0.294\ J} \]

(b) Charging from \(V_1\) to \(V_2\) with a supply \(V_2\): the supply delivers \(C(V_2 - V_1)V_2\), storage rises by \(\tfrac12C(V_2^2 - V_1^2)\), so the heat is \(\tfrac12C(V_2 - V_1)^2\). Each 12.5 V step:

\[ \tfrac12(470\times10^{-6})(12.5)^2 = 0.0367\ \text{J},\qquad \text{two steps: } \mathbf{0.073\ J} \]

Half the loss of a single step. With \(N\) steps the loss falls as \(1/N\) — the idea behind “adiabatic” low-power logic and stepped-voltage capacitor drivers.

10Capacitor at t = 0 and t → ∞10 pts

A 12 V battery is in series with \(R_1 = 4\ \Omega\), which feeds two parallel branches: \(R_2 = 8\ \Omega\) alone, and \(R_3 = 8\ \Omega\) in series with an uncharged \(C = 50\ \mu\text{F}\). The switch closes at \(t = 0\). Find (a) the battery current and the capacitor-branch current just after closing; (b) the battery current, \(V_C\) and the capacitor charge a long time later; (c) the time constant of the transient.

Solution

(a) Uncharged capacitor = wire. Branches are \(8\parallel8 = 4\ \Omega\):

\[ I(0^+) = \frac{12}{4 + 4} = \mathbf{1.5\ A},\qquad \text{split equally: } I_C(0^+) = \mathbf{0.75\ A} \]

(b) Charged capacitor = open circuit; no current in \(R_3\), so \(V_C = V_{R_2}\):

\[ I(\infty) = \frac{12}{4 + 8} = \mathbf{1.0\ A},\qquad V_C = 1.0(8) = \mathbf{8.0\ V},\qquad Q = CV_C = (50\times10^{-6})(8.0) = \mathbf{400\ \mu C} \]

(c) Replace the battery by a wire and look back from the capacitor: \(R_3\) in series with \(R_1\parallel R_2\):

\[ R_{\text{th}} = 8 + \frac{4\cdot8}{12} = 10.67\ \Omega,\qquad \tau = R_{\text{th}}C = (10.67)(50\times10^{-6}) = \mathbf{0.53\ ms} \]

Using \(R_3\) alone (0.40 ms) or the “total” resistance is the classic error.

Scoring

Score	Where you are
90–100	Solid. On to magnetism.
75–89	Method fine, slips in signs or bookkeeping. Re-check the sign table in Section 6 and always finish with a power balance.
60–74	RC trouble (Q7–Q10)? Re-read Section 8 with the lab open. Meter or source trouble (Q3, Q6)? Section 4 and Section 7.
< 60	Rebuild from Section 1 through Section 5: \(I = nqv_dA\), \(R = \rho L/A\), \(P = IV\) and series/parallel must be automatic first.

Where the points usually go

Using the radius as the diameter in \(A = \pi r^2\) (Q1, Q2); forgetting that a cord or circuit path goes out and back; flipping a negative current mid-solution (see the two-loop example in Unit 5, Section 6); putting a shunt in series (Q6); using the wrong resistance in \(\tau\) (Q10); and expecting a smaller \(R\) to waste less energy while charging (Q9).