The Workbook

1. \(I = \dfrac{E}{R + r} = \dfrac{12}{5 + 1} = 2\) A.

2. \(V = IR = 2 \times 5 = 10\) V. Drop inside the cell \(= Ir = 2 \times 1 = 2\) V. Check: \(10 + 2 = 12\) V.

The emf is the work done by the cell per unit charge moved through it. It equals the potential difference across the terminals when no current flows.

The terminal voltage is the potential difference across the terminals when the cell is in a circuit: \(V = E - Ir\) when discharging.

\(V = E\) when \(I = 0\) (open circuit, or \(r = 0\)). \(V = E + Ir\), greater than \(E\), when the cell is being charged.

With no current \(V = E = 2\) V. With \(1\) A: \(1.5 = 2 - 1 \times r\), so \(r = 0.5\ \Omega\).

\(R = \dfrac{V}{I} = \dfrac{1.5}{1} = 1.5\ \Omega\).

1. \(V = E - Ir\). At \(I = 0\), \(V = E = 6\) V.

2. The slope is \(-r\). \(V\) falls from \(6\) to \(0\) as \(I\) rises from \(0\) to \(3\) A: \(r = \dfrac{6}{3} = 2\ \Omega\).

3. Short circuit means \(V = 0\): \(I = \dfrac{E}{r} = 3\) A, where the line meets the \(I\) axis.

1. \(I = \dfrac{6}{4 + 2} = 1\) A. \(V = IR = 4\) V.

2. \(P_R = I^2R = 4\) W. \(P_r = I^2r = 2\) W. Total from the cell: \(EI = 6\) W.

3. Efficiency \(= \dfrac{P_R}{EI} = \dfrac{4}{6} = \dfrac{2}{3}\), the same as \(\dfrac{R}{R + r} = \dfrac{V}{E}\).

1. \(I = \dfrac{E}{R + r}\), so \(P = I^2R = \dfrac{E^2R}{(R + r)^2}\).

2. Write \((R + r)^2 = (R - r)^2 + 4Rr\). Then \(P = \dfrac{E^2R}{(R - r)^2 + 4Rr} = \dfrac{E^2}{(R - r)^2/R + 4r}\). The denominator is smallest when \(R = r\), so \(P\) is largest there: \(P_{\max} = \dfrac{E^2}{4r}\).

1. The peak is at \(R = 2\ \Omega\), so \(r = 2\ \Omega\).

2. \(P_{\max} = 4.5\) W \(= \dfrac{E^2}{4r} = \dfrac{E^2}{8}\), so \(E^2 = 36\) and \(E = 6\) V.

3. \(P = \dfrac{E^2R}{(R + r)^2} = \dfrac{36 \times 6}{64} = 3.375\) W.

\(P = \dfrac{E^2R}{(R + r)^2}\). For \(R = \dfrac{r}{2}\): \(\dfrac{E^2 (r/2)}{(3r/2)^2} = \dfrac{2}{9}\dfrac{E^2}{r}\). For \(R = r\): \(\dfrac{1}{4}\dfrac{E^2}{r}\). For \(R = 2r\): \(\dfrac{2E^2 r}{9r^2} = \dfrac{2}{9}\dfrac{E^2}{r}\).

The loads \(\dfrac{r}{2}\) and \(2r\) receive the same power. In general, \(R\) and \(\dfrac{r^2}{R}\) give equal power.

1. During charging the current is forced in against the emf: \(V = E + Ir = 12 + 4 \times 0.5 = 14\) V.

2. Stored: \(EI = 12 \times 4 = 48\) W. Heat: \(I^2r = 16 \times 0.5 = 8\) W. The charger supplies \(VI = 14 \times 4 = 56\) W \(= 48 + 8\).

\(EI\) is the power supplied by the chemical reaction. \(VI\) is the power delivered to the outside circuit and \(I^2r\) is the heat inside the cell.

Energy is conserved: \(EI = VI + I^2r\). Divide by \(I\): \(E = V + Ir\), so \(V = E - Ir\).

The starter motor draws a very large current, perhaps \(100\) A. The drop \(Ir\) inside the battery becomes large, so the terminal voltage \(V = E - Ir\) falls. The headlights are across the terminals, so they get less voltage and dim. When the motor stops, \(I\) falls and \(V\) recovers.

1. Net emf \(= 4 - 2 = 2\) V. Total resistance \(= 1 + 1 + 2 = 4\ \Omega\). \(I = \dfrac{2}{4} = 0.5\) A, driven by the \(4\) V cell.

2. The \(4\) V cell discharges: \(V = 4 - 0.5 \times 1 = 3.5\) V. The \(2\) V cell is being charged: \(V = 2 + 0.5 \times 1 = 2.5\) V. Check: \(3.5 - 2.5 = 1\) V \(= 0.5 \times 2\) across the resistor.

Series: emfs add, \(E = 4 \times 2 = 8\) V; internal resistances add, \(r = 4 \times 1 = 4\ \Omega\).

\(I = \dfrac{8}{4 + 4} = 1\) A.

Parallel identical cells: emf stays \(2\) V; internal resistance \(= \dfrac{1}{4} = 0.25\ \Omega\).

\(I = \dfrac{2}{0.75 + 0.25} = 2\) A.

Series gives \(I = \dfrac{nE}{R + nr}\), parallel gives \(I = \dfrac{E}{R + r/n}\).

If \(R \gg r\) the internal resistance hardly matters and series wins because the emf is \(n\) times larger. If \(R \ll r\) the internal resistance controls the current and parallel wins because it is \(n\) times smaller.

1. Each row: \(4 \times 2 = 8\) V and \(4\ \Omega\). Three rows in parallel: \(E = 8\) V, \(r = \dfrac{4}{3}\ \Omega\).

2. \(I = \dfrac{8}{4/3 + 4/3} = \dfrac{8}{8/3} = 3\) A.

3. For \(m\) rows of \(n\) cells the internal resistance is \(\dfrac{nr}{m}\). The current is largest when this equals \(R\) (Question 6). Here \(\dfrac{nr}{m} = \dfrac{4}{3} = R\), so the current is the largest possible.

\(E_{\text{eq}} = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \dfrac{6 + 3}{2} = 4.5\) V. \(r_{\text{eq}} = \dfrac{r_1 r_2}{r_1 + r_2} = 0.5\ \Omega\).

\(I = \dfrac{4.5}{4 + 0.5} = 1\) A.

The reversed cell subtracts its emf instead of adding it: net emf \(= (n - 1)E - E = (n - 2)E\). Internal resistances add regardless of direction: \(nr\).

For \(n = 6\): \(4E\) and \(6r\).

1. Switch open: no current, so the voltmeter reads the emf, \(E = 6\) V. Switch closed: \(5 = 6 - 2r\), so \(r = 0.5\ \Omega\).

2. \(R = \dfrac{V}{I} = \dfrac{5}{2} = 2.5\ \Omega\).

Load: \(\dfrac{4 \times 12}{16} = 3\ \Omega\). \(I = \dfrac{12}{3 + 1} = 3\) A. Terminal voltage \(V = 3 \times 3 = 9\) V.

Currents: \(\dfrac{9}{4} = 2.25\) A in the \(4\ \Omega\) and \(\dfrac{9}{12} = 0.75\) A in the \(12\ \Omega\). Check: \(2.25 + 0.75 = 3\) A.

\(E = 3\) V, \(r = 1\ \Omega\). \(I = \dfrac{3}{2 + 1} = 1\) A. Bulb: \(I^2R = 2\) W. Cells: \(I^2r = 1\) W. Total \(3\) W \(= EI\).

\(5.5 = E - r\) and \(5 = E - 2r\). Subtract: \(0.5 = r\), so \(r = 0.5\ \Omega\) and \(E = 6\) V.

Balance: \(\dfrac{P}{Q} = \dfrac{R}{S}\). The two arms \(P\) and \(R\) then drop the same fraction of the cell voltage. So the junctions \(B\) and \(D\) are at the same potential. With no potential difference across the galvanometer, no current flows through it.

No. At balance the galvanometer carries no current, so removing it leaves every current exactly as it was. The two branches \(P\)-\(Q\) and \(R\)-\(S\) behave as two independent series pairs across the cell.

1. \(\dfrac{X}{R} = \dfrac{l}{100 - l} = \dfrac{40}{60}\), so \(X = 6 \times \dfrac{2}{3} = 4\) Ω.

2. With \(R\) on the left: \(\dfrac{6}{4} = \dfrac{l}{100 - l}\), so \(l = 60\) cm. The null point mirrors about the middle.

The error in \(X\) from a small error \(\delta l\) in reading the null point is least when \(l = 50\) cm, because \(\dfrac{\delta X}{X} = \delta l\left(\dfrac{1}{l} + \dfrac{1}{100 - l}\right)\) is smallest there. The bridge is also most sensitive when all four arms are of the same order.

First: \(\dfrac{2}{3} = \dfrac{40.1 + \alpha}{59.9 + \beta}\), so \(2(59.9 + \beta) = 3(40.1 + \alpha)\): \(3\alpha - 2\beta = -0.5\). Second: \(\dfrac{3}{2} = \dfrac{60.4 + \alpha}{39.6 + \beta}\), so \(3(39.6 + \beta) = 2(60.4 + \alpha)\): \(2\alpha - 3\beta = -2\). Solving: \(\alpha = 0.5\) cm and \(\beta = 1\) cm.

No. At the null point no current flows through the galvanometer branch, so its resistance does not matter. While hunting for the null point, the series resistance keeps the galvanometer current small and protects the instrument. It is shorted near balance for full sensitivity.

Uniform warming multiplies the resistance of every centimetre of the wire by the same factor, so the ratio \(\dfrac{l}{100 - l}\) at balance is unchanged: no shift. If \(X\) alone warms, its resistance rises, so the null point moves to the right (larger \(l\)) to keep \(\dfrac{X}{R} = \dfrac{l}{100 - l}\).

Yes. The condition \(\dfrac{P}{Q} = \dfrac{R}{S}\) can be rewritten \(\dfrac{P}{R} = \dfrac{Q}{S}\), which is the balance condition for the bridge with the cell and galvanometer swapped. Balance does not depend on which diagonal carries the cell.

1. \(X = R\dfrac{Q}{P} = R\). The reversal shows \(X\) lies between \(27\) and \(28\) Ω.

2. \(X = \dfrac{R}{10}\), so \(X\) lies between \(27.3\) and \(27.4\) Ω.

3. \(X = \dfrac{R}{100} = \dfrac{2734}{100} = 27.34\) Ω.

\(X = R\dfrac{Q}{P}\). With \(P = 10\), \(100\), \(1000\) and \(Q = 10\), the box reads \(X\) to \(1\) Ω, then \(0.1\) Ω, then \(0.01\) Ω. The same resistance box in steps of \(1\) Ω serves every stage. Each step adds one decimal place.

The coils have some inductance, so the current takes a moment to settle when the battery key is closed. If the galvanometer were already connected, it would receive a transient jolt that has nothing to do with balance. Pressing the battery key first lets the current become steady.

1. The whole \(2\) V drops across the wire: gradient \(= \dfrac{2}{4} = 0.5\) V/m.

2. \(E = 0.5 \times 2.4 = 1.2\) V.

3. Current \(= \dfrac{2}{8 + 8} = 0.125\) A, drop across the wire \(= 1\) V, gradient \(0.25\) V/m. Balancing \(1.2\) V would need \(4.8\) m, longer than the wire, so no null point is found.

1. With \(K_2\) closed the cell drives a current through \(R\). The potentiometer now balances the terminal voltage \(V = E - Ir\), which is less than \(E\). A smaller voltage needs a shorter length.

2. \(\dfrac{E}{V} = \dfrac{l_1}{l_2}\) and \(\dfrac{E}{V} = \dfrac{R + r}{R}\), so \(r = R\dfrac{l_1 - l_2}{l_2} = 10 \times \dfrac{0.5}{2.5} = 2\) Ω.

At balance \(E = kl\) with the same gradient \(k\): \(\dfrac{E_1}{E_2} = \dfrac{3}{2}\).

At the null point the potentiometer draws no current from the cell. There is no drop across the internal resistance, so the balanced voltage is the full emf. A voltmeter must draw a small current \(I\) to deflect, so it reads \(V = E - Ir\).

(i) The gradient rises, so the same \(E\) balances at a shorter \(l\). (ii) The current falls, the gradient falls, and \(l\) increases. (iii) No point on the wire has a potential drop as large as \(E\). The galvanometer deflects the same way everywhere: no null point.

\(I = \dfrac{2}{4 + 4} = 0.25\) A. Drop across the wire \(= 0.25 \times 4 = 1\) V. Gradient \(= \dfrac{1}{4} = 0.25\) V/m.

Reduce the potential gradient: use a longer wire, or add resistance in series with the wire to lower the current. A smaller gradient means a given change in \(E\) moves the null point through a larger length, which is easier to read.

1. \(\dfrac{P}{Q} = \dfrac{2}{3}\) but \(\dfrac{R}{S} = \dfrac{4}{8} = \dfrac{1}{2}\). Not balanced.

2. Branch \(ABC\): \(I = \dfrac{10}{5} = 2\) A, so \(V_B = 3 \times 2 = 6\) V above \(C\). Branch \(ADC\): \(I = \dfrac{10}{12}\) A, so \(V_D = 8 \times \dfrac{10}{12} \approx 6.67\) V.

3. \(D\) is higher, so current flows from \(D\) to \(B\) through \(G\). Balance needs \(\dfrac{4}{S} = \dfrac{2}{3}\), so \(S = 6\) Ω.

Balance needs \(R = S\) when \(P = Q\). Either raise \(R\) to \(5.2\) Ω or lower \(S\) to \(5\) Ω; in practice the adjustable arm is changed by \(0.2\) Ω.

\(\dfrac{X}{5} = \dfrac{50}{50}\), so \(X = 5\) Ω. With \(5\) Ω in parallel the left gap has \(2.5\) Ω: \(\dfrac{2.5}{5} = \dfrac{l}{100 - l}\), so \(100 - l = 2l\) and \(l = 33.3\) cm.

\(E \propto l\): \(E_2 = 1.5 \times \dfrac{250}{300} = 1.25\) V.

The strips join the wire ends to the gaps holding \(X\) and \(R\). Thick copper has negligible resistance. The only resistances in the bridge arms are then \(X\), \(R\) and the two parts of the wire, as the formula assumes.

Open circuit: \(E = kl_1\). With \(R\) across the cell the current is \(I = \dfrac{E}{R + r}\) and the terminal voltage \(V = IR = \dfrac{ER}{R + r} = kl_2\). Dividing: \(\dfrac{E}{V} = \dfrac{R + r}{R} = \dfrac{l_1}{l_2}\), so \(1 + \dfrac{r}{R} = \dfrac{l_1}{l_2}\) and \(r = R\dfrac{l_1 - l_2}{l_2}\).

Current sensitivity is the deflection per unit current, \(\dfrac{\theta}{I}\). Voltage sensitivity is the deflection per unit voltage, \(\dfrac{\theta}{V} = \dfrac{\theta}{IG}\). Here \(\dfrac{\theta}{I} = 20\) div/mA and \(\dfrac{\theta}{V} = \dfrac{20}{1 \times 50} = 0.4\) div/mV.

The coil is fine and deflects fully at a current of only \(I_g\), typically a milliampere or less. More current would drive the needle past the stop and burn the coil. Likewise it needs only \(I_gG\), a fraction of a volt, for full deflection, so a few volts would overload it.

1. \(G\) and \(S\) are in parallel, so they have the same potential difference: \(I_gG = (I - I_g)S\).

2. \(S = \dfrac{I_gG}{I - I_g} = \dfrac{10^{-3} \times 100}{0.999} \approx 0.1\) Ω.

3. \(R_A = \dfrac{GS}{G + S} = \dfrac{100 \times 0.1}{100.1} \approx 0.1\) Ω.

The shunt must carry almost all the current, leaving only \(I_g\) for the coil, so it must be much smaller than \(G\). Only a parallel connection divides the current between two paths while keeping the same voltage across both.

1. At full scale \(I_g\) flows through \(G + R\) with \(10\) V across: \(R = \dfrac{V}{I_g} - G = \dfrac{10}{10^{-3}} - 100 = 9900\) Ω.

2. \(R_V = G + R = 10^{4}\) Ω.

\(S = \dfrac{10^{-3} \times 100}{5 - 0.001} = \dfrac{0.1}{4.999} \approx 0.02\) Ω. A larger range needs a smaller shunt.

Half scale. The current divides in the fixed ratio \(S : G\), so the coil current is always the same fraction \(\dfrac{S}{S + G}\) of the total. Half the total gives half of \(I_g\), and the deflection is proportional to the coil current.

\(\dfrac{I_g}{I} = \dfrac{10^{-3}}{1} = 0.1\%\); the shunt carries \(99.9\%\). An ammeter goes in series with the circuit element, so its resistance adds to the circuit. A low resistance keeps the current almost what it was before the meter was inserted.

\(S = \dfrac{I_gG}{I - I_g} = \dfrac{0.002 \times 50}{2 - 0.002} = \dfrac{0.1}{1.998} \approx 0.05\) Ω.

1. \(I = \dfrac{12}{5 + 1} = 2\) A.

2. Without the meter: \(\dfrac{12}{5} = 2.4\) A.

3. Error \(= \dfrac{2.4 - 2}{2.4} = 16.7\%\). The meter's own resistance reduces the current it is trying to measure.

1. \(1000\) Ω in parallel with \(1000\) Ω: \(500\) Ω.

2. Total resistance \(1500\) Ω, current \(\dfrac{12}{1500} = 8\) mA. Reading \(= 0.008 \times 500 = 4\) V.

3. With no current drawn, the two equal resistors share \(12\) V equally: \(6\) V.

A voltmeter goes across the element. If its resistance is high it draws almost no current, so the currents and voltages in the circuit stay as they were. An ideal voltmeter has infinite resistance; an ideal ammeter has zero resistance.

\(1000 \times 10 = 10^{4}\) Ω. Across the \(1000\) Ω resistor: parallel value \(\dfrac{1000 \times 10000}{11000} \approx 909\) Ω, current \(\dfrac{12}{1909} \approx 6.3\) mA, reading \(\approx 5.7\) V. Much closer to the true \(6\) V than the \(4\) V of Q11, because the meter draws far less current.

\(R = \dfrac{V}{I_g} - G = \dfrac{5}{0.002} - 50 = 2500 - 50 = 2450\) Ω.

For \(20\) V: \(R' = \dfrac{20}{0.002} - 50 = 9950\) Ω. Extra resistance \(= 9950 - 2450 = 7500\) Ω.

Total resistance \(100 + 9900 = 10^{4}\) Ω. Current \(= \dfrac{5}{10^{4}} = 0.5\) mA, half of \(I_g\): the needle stands at half scale. This is just the \(10\) V voltmeter of Q5 reading \(5\) V.

1. \(R_1 = \dfrac{10}{10^{-3}} - 100 = 9900\) Ω.

2. \(R_1 + R_2 = \dfrac{100}{10^{-3}} - 100 = 99900\) Ω, so \(R_2 = 90000\) Ω.

The added \(0.1\) Ω equals the meter's resistance, so at full deflection it carries the same \(1\) A as the meter. Total \(= 2\) A.

An ammeter has a very low resistance, about \(0.1\) Ω or less. Across a cell it lets a current of many amperes flow, limited only by the cell's internal resistance, far above the meter's full-scale value. The shunt or coil overheats and may burn.

The ammeter goes in series so that the whole current passes through it; its low resistance then hardly changes that current. The voltmeter goes in parallel so that it has the element's voltage across it; its high resistance then draws almost no current.

Full scale is \(20\) divisions for \(1\) A, so one division is \(\dfrac{1}{20}\) A \(= 50\) mA. The shunt trades sensitivity for range.

\(\dfrac{I_g}{I} = \dfrac{S}{S + G} = \dfrac{1}{101} \approx 1\%\). The rest, about \(99\%\), passes through the shunt.

\(R_A = \dfrac{GS}{G + S} = \dfrac{S}{1 + S/G}\). When \(S \ll G\) the denominator is nearly \(1\), so \(R_A \approx S\). For Q3, \(\dfrac{100 \times 0.1}{100.1} = 0.0999\) Ω.

The coil carries \(I_g\) and the shunt carries the rest, \(I - I_g\). Equal potential differences: \(I_gG = (I - I_g)S\). Solving for the shunt: \(S = \dfrac{I_gG}{I - I_g}\).