The Timpson Lyceum

1. (1 pt •) Three resistors, $R_1=4\,\Omega$ , $R_2=12\,\Omega$ , and $R_3=6\,\Omega$ , are available.

(a) Find their equivalent resistance if all three are wired in series.

(b) Find their equivalent resistance if all three are wired in parallel.

Solution

1.

(a) Series adds directly: $R_\text{series}=4+12+6=22\,\Omega$ .

(b) Parallel adds reciprocals: $\dfrac{1}{R_\text{par}}=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{3+1+2}{12}=\dfrac{6}{12}=\dfrac12$ , so $R_\text{par}=2\,\Omega$ .

2. (2 pts ••) A $24\,\text{V}$ battery (assume no internal resistance) is connected to a resistor $R_1=6\,\Omega$ in series with a parallel combination of $R_2=12\,\Omega$ and $R_3=6\,\Omega$ .

(a) Find the equivalent resistance of the whole network and the total current the battery delivers.

(b) Find the voltage across the parallel combination.

(c) Find the current through $R_2$ and through $R_3$ .

Solution

2.

(a) The parallel pair: $\dfrac{1}{R_p}=\dfrac{1}{12}+\dfrac16=\dfrac{1+2}{12}=\dfrac14$ , so $R_p=4\,\Omega$ . In series with $R_1$ : $R_\text{eq}=6+4=10\,\Omega$ . Total current $I=\dfrac{24}{10}=2.4\,\text{A}$ .

(b) All the current passes through $R_1$ , dropping $IR_1=(2.4)(6)=14.4\,\text{V}$ . The parallel section therefore has $V_p=24-14.4=9.6\,\text{V}$ (equivalently $I R_p=2.4\times4=9.6\,\text{V}$ ).

(c) $I_2=\dfrac{V_p}{R_2}=\dfrac{9.6}{12}=0.80\,\text{A}$ and $I_3=\dfrac{V_p}{R_3}=\dfrac{9.6}{6}=1.6\,\text{A}$ . Check: $0.80+1.6=2.4\,\text{A}$ . ✓

3. (2 pts ••) A battery of emf $\mathcal{E}=12\,\text{V}$ has internal resistance $r=0.5\,\Omega$ and is connected to an external resistor $R=5.5\,\Omega$ .

(a) Find the current in the circuit.

(b) Find the terminal voltage of the battery.

(c) Find the power dissipated in $R$ and the power dissipated inside the battery.

Solution

3.

(a) $I=\dfrac{\mathcal{E}}{R+r}=\dfrac{12}{5.5+0.5}=\dfrac{12}{6}=2.0\,\text{A}$ .

(b) Terminal voltage $V=\mathcal{E}-Ir=12-(2.0)(0.5)=11\,\text{V}$ (equivalently $IR=2.0\times5.5=11\,\text{V}$ ).

(c) $P_R=I^2R=(2.0)^2(5.5)=22\,\text{W}$ ; internal loss $P_r=I^2r=(2.0)^2(0.5)=2.0\,\text{W}$ . (The battery converts $\mathcal{E}I=24\,\text{W}$ total, $22\,\text{W}$ to the load, $2\,\text{W}$ wasted internally.)

4. (2 pts ••) An ammeter and a voltmeter are idealized measuring instruments.

(a) An ideal ammeter measures the current through a branch. Should it have very large or very small resistance, and should it be connected in series or in parallel with that branch? Explain what would go wrong if its resistance were not ideal.

(b) An ideal voltmeter measures the potential difference across an element. Should it have very large or very small resistance, and should it be connected in series or in parallel with that element? Explain what would go wrong if its resistance were not ideal.

Solution

4.

(a) An ideal ammeter has very small (ideally zero) resistance and is connected in series with the branch, so the same current flows through it as through the branch. If its resistance were appreciable, inserting it would add resistance to the branch, reducing the very current it is trying to measure and disturbing the circuit.

(b) An ideal voltmeter has very large (ideally infinite) resistance and is connected in parallel with the element. It must draw negligible current so it does not divert current from the element; if its resistance were too small it would act as a parallel path, reducing the potential difference it is trying to measure (loading the circuit).

5. (2 pts ••) A capacitor $C=50\,\mu\text{F}$ , initially uncharged, is connected in series with a resistor $R=100\,\text{k}\Omega$ and a $\mathcal{E}=9\,\text{V}$ battery through a switch that closes at $t=0$ .

(a) Find the time constant $\tau$ and the initial charging current $i(0)$ .

(b) Find the final charge on the capacitor once it is fully charged.

(c) Find the charge on the capacitor and the current in the circuit at $t=10\,\text{s}$ .

(d) How long after closing the switch does the charge reach $90\%$ of its final value?

Solution

5.

(a) $\tau=RC=(100\times10^{3})(50\times10^{-6})=5.0\,\text{s}$ . Initial current (capacitor acts like a wire at $t=0$ ): $i(0)=\dfrac{\mathcal{E}}{R}=\dfrac{9}{100\times10^{3}}=9.0\times10^{-5}\,\text{A}=90\,\mu\text{A}$ .

(b) Final charge $Q_f=C\mathcal{E}=(50\times10^{-6})(9)=4.5\times10^{-4}\,\text{C}=450\,\mu\text{C}$ .

(c) At $t=10\,\text{s}$ , $t/\tau=2$ , so $e^{-2}=0.135$ .

$q=Q_f(1-e^{-2})=450\,\mu\text{C}\,(1-0.135)=450(0.865)\approx 3.9\times10^{2}\,\mu\text{C}=390\,\mu\text{C}$ .

$i=i(0)\,e^{-2}=90\,\mu\text{A}\,(0.135)\approx 12\,\mu\text{A}$ .

(d) Set $1-e^{-t/\tau}=0.90\Rightarrow e^{-t/\tau}=0.10\Rightarrow t=-\tau\ln(0.10)=5.0\times\ln10=5.0(2.303)\approx 12\,\text{s}$ .

6. (2 pts ••) A capacitor $C=100\,\mu\text{F}$ is charged to $V_0=12\,\text{V}$ and then, at $t=0$ , connected across a resistor $R=20\,\text{k}\Omega$ (the battery is removed) so it discharges.

(a) Find the time constant and the initial discharge current.

(b) Find the voltage across the capacitor at $t=3\,\text{s}$ .

(c) Find the energy stored in the capacitor at $t=0$ . Where does this energy go as the capacitor discharges?

Solution

6.

(a) $\tau=RC=(20\times10^{3})(100\times10^{-6})=2.0\,\text{s}$ . Initial discharge current $i(0)=\dfrac{V_0}{R}=\dfrac{12}{20\times10^{3}}=6.0\times10^{-4}\,\text{A}=0.60\,\text{mA}$ .

(b) Discharging: $V(t)=V_0e^{-t/\tau}$ . At $t=3\,\text{s}$ , $t/\tau=1.5$ , $e^{-1.5}=0.223$ , so $V=12(0.223)\approx 2.7\,\text{V}$ .

(c) $U_0=\tfrac12 CV_0^2=\tfrac12(100\times10^{-6})(12)^2=\tfrac12(100\times10^{-6})(144)=7.2\times10^{-3}\,\text{J}=7.2\,\text{mJ}$ . As the capacitor discharges, all of this stored energy is dissipated as heat ( $i^2R$ ) in the resistor.

7. (1 pt •) In the charging circuit of Problem 5, at the instant the switch closes ( $t=0$ ) the capacitor is uncharged.

(a) What is the potential difference across the capacitor at that instant, and what is the potential difference across the resistor?

(b) After a very long time, what is the current in the circuit, and what fraction of the battery's total energy output has been stored in the capacitor versus dissipated in the resistor?

Solution

7.

(a) At $t=0$ the uncharged capacitor has $V_C=q/C=0$ , so it behaves like a short: the full emf appears across the resistor, $V_R=\mathcal{E}=9\,\text{V}$ .

(b) After a long time the capacitor is fully charged, no current flows: $i(\infty)=0$ . The battery does total work $W=Q_f\mathcal{E}=C\mathcal{E}^2$ ; of this, $\tfrac12 C\mathcal{E}^2$ ends up stored in the capacitor and $\tfrac12 C\mathcal{E}^2$ is dissipated in the resistor — a 50/50 split, independent of $R$ (proved in the Extra Credit).

8. (3 pts •••) Two batteries and three resistors form the following two-loop network. The left branch contains a battery $\mathcal{E}_1=12\,\text{V}$ in series with $R_1=4\,\Omega$ ; the middle branch contains a battery $\mathcal{E}_2=9\,\text{V}$ in series with $R_2=3\,\Omega$ ; the right branch contains only $R_3=6\,\Omega$ . All three branches run between the same top node $a$ and bottom node $b$ , with each battery's $+$ terminal toward node $a$ .

(a) Define a current in each branch and write Kirchhoff's junction rule at node $a$ .

(b) Write two independent loop equations and solve for all three branch currents. State the direction of each current (toward or away from node $a$ ).

(c) For each battery, determine whether it is delivering energy or absorbing energy (being charged), based on whether its current leaves or enters its $+$ terminal. Then verify overall energy balance: show that the total power delivered by the sources equals the total power dissipated in the three resistors.

Solution

8.

(a) Let $I_1$ be the current in the left branch flowing toward node $a$ , $I_2$ the current in the middle branch flowing toward $a$ , and $I_3$ the current in the right branch flowing away from $a$ (down through $R_3$ ). Junction rule at $a$ :

I_1+I_2=I_3.

(b) Let $V\equiv V_a-V_b$ be the potential of the top node above the bottom node. Traversing each branch from $b$ up to $a$ :

Left branch: $V=\mathcal{E}_1-I_1R_1\Rightarrow I_1=\dfrac{\mathcal{E}_1-V}{R_1}=\dfrac{12-V}{4}$ .

Middle branch: $V=\mathcal{E}_2-I_2R_2\Rightarrow I_2=\dfrac{\mathcal{E}_2-V}{R_2}=\dfrac{9-V}{3}$ .

Right branch: $V=I_3R_3\Rightarrow I_3=\dfrac{V}{6}$ .

(These two branch relations together with $V=I_3R_3$ are equivalent to the two independent loop equations for the left–right loop and the middle–right loop.) Substituting into the junction rule:

\frac{12-V}{4}+\frac{9-V}{3}=\frac{V}{6}.

Multiply through by 12: $3(12-V)+4(9-V)=2V$ , i.e. $36-3V+36-4V=2V$ , so $72=9V$ and $V=8.0\,\text{V}$ .

Then $I_1=\dfrac{12-8}{4}=1.0\,\text{A}$ (toward $a$ ), $I_2=\dfrac{9-8}{3}=0.33\,\text{A}$ (toward $a$ ), $I_3=\dfrac{8}{6}=1.33\,\text{A}$ (away from $a$ , down through $R_3$ ). Check: $1.0+0.33=1.33\,\text{A}$ . ✓ All currents came out positive, so every assumed direction is correct.

(c) Each current came out positive, meaning it leaves the battery at the $+$ terminal (toward node $a$ ). So both batteries are delivering energy:

$\mathcal{E}_1$ : $I_1=1.0\,\text{A}$ leaves its $+$ terminal → delivering $\mathcal{E}_1 I_1=12(1.0)=12.0\,\text{W}$ .

$\mathcal{E}_2$ : $I_2=0.33\,\text{A}$ leaves its $+$ terminal → delivering $\mathcal{E}_2 I_2=9(0.333)=3.0\,\text{W}$ .

Neither is being charged here. (A battery absorbs only when its branch current is forced into its $+$ terminal, i.e. when the rest of the network pushes a current opposite to what that battery "wants" — which would show up as a negative branch current with the assumed directions.) Energy balance:

Power delivered $=\mathcal{E}_1 I_1+\mathcal{E}_2 I_2=12.0+3.0=15.0\,\text{W}$ .

Power dissipated $=I_1^2R_1+I_2^2R_2+I_3^2R_3=(1.0)^2(4)+(0.333)^2(3)+(1.333)^2(6)=4.0+0.33+10.67=15.0\,\text{W}$ . ✓

Extra Credit (3 pts •••)

Derive the charging law for the series R–C circuit of Problem 5 from first principles.

(a) Apply Kirchhoff's loop rule at an arbitrary time $t$ to write the first-order differential equation governing the charge $q(t)$ on the capacitor, using $i=dq/dt$ .

(b) Solve the differential equation by separation of variables with the initial condition $q(0)=0$ , and show that

q(t)=C\mathcal{E}\left(1-e^{-t/RC}\right).

(c) Differentiate to obtain $i(t)$ , and prove by direct integration that the total energy delivered by the battery over all time, $\int_0^{\infty}\mathcal{E}\,i\,dt$ , splits into exactly $\tfrac12 C\mathcal{E}^2$ stored in the capacitor and $\tfrac12 C\mathcal{E}^2$ dissipated in the resistor — independent of the value of $R$ .

Solution

Extra Credit

(a) At time $t$ the capacitor holds charge $q$ , so its voltage is $q/C$ ; the resistor carries current $i=dq/dt$ and drops $iR$ . Kirchhoff's loop rule around the single loop (battery → resistor → capacitor):

\mathcal{E}-iR-\frac{q}{C}=0 \quad\Longrightarrow\quad R\frac{dq}{dt}=\mathcal{E}-\frac{q}{C}.

(b) Rearrange and separate variables:

\frac{dq}{\mathcal{E}-q/C}=\frac{dt}{R} \quad\Longrightarrow\quad \frac{C\,dq}{C\mathcal{E}-q}=\frac{dt}{R}.

Integrate from $0$ to $t$ (with $q(0)=0$ ):

-C\ln\!\left(C\mathcal{E}-q\right)\Big|_{0}^{q}=\frac{t}{R},

so $-C\big[\ln(C\mathcal{E}-q)-\ln(C\mathcal{E})\big]=\dfrac{t}{R}$ , i.e.

\ln\!\frac{C\mathcal{E}-q}{C\mathcal{E}}=-\frac{t}{RC}.

Exponentiate: $C\mathcal{E}-q=C\mathcal{E}\,e^{-t/RC}$ , hence

q(t)=C\mathcal{E}\left(1-e^{-t/RC}\right).

As $t\to\infty$ , $q\to C\mathcal{E}=Q_f$ and $q(0)=0$ , as required.

(c) Differentiate:

i(t)=\frac{dq}{dt}=C\mathcal{E}\cdot\frac{1}{RC}e^{-t/RC}=\frac{\mathcal{E}}{R}\,e^{-t/RC}.

Total energy delivered by the battery:

W_\text{batt}=\int_0^{\infty}\mathcal{E}\,i\,dt=\frac{\mathcal{E}^2}{R}\int_0^{\infty}e^{-t/RC}\,dt=\frac{\mathcal{E}^2}{R}\big[-RC\,e^{-t/RC}\big]_0^{\infty}=\frac{\mathcal{E}^2}{R}(RC)=C\mathcal{E}^2.

Energy dissipated in the resistor:

W_R=\int_0^{\infty}i^2R\,dt=R\cdot\frac{\mathcal{E}^2}{R^2}\int_0^{\infty}e^{-2t/RC}\,dt=\frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2}=\frac12 C\mathcal{E}^2.

Energy stored in the capacitor at the end:

W_C=\frac{Q_f^2}{2C}=\frac{(C\mathcal{E})^2}{2C}=\frac12 C\mathcal{E}^2.

Indeed $W_C+W_R=\tfrac12 C\mathcal{E}^2+\tfrac12 C\mathcal{E}^2=C\mathcal{E}^2=W_\text{batt}$ . Exactly half the battery's energy is stored and half is dissipated, and since the $R$ 's canceled, this split is independent of $R$ (and of $C$ ). $\blacksquare$

Chapter 26 — Direct-Current Circuits

Chapter 26 — Direct-Current Circuits

Extra Credit (3 pts •••)

Extra Credit