The Workbook

1. \(q = It = 3.2 \times 10 = 32\) C.

2. Charge per second \(= 3.2\) C. Number of electrons \(= \dfrac{3.2}{1.6 \times 10^{-19}} = 2 \times 10^{19}\).

1. \(I = \dfrac{q}{t} = \dfrac{600}{300} = 2\) A.

2. \(t = \dfrac{q}{I} = \dfrac{1}{2} = 0.5\) s.

1. \(q = \displaystyle\int_0^2 (2t + 3)\,dt = [t^2 + 3t]_0^2 = 4 + 6 = 10\) C.

2. \(I_{\text{av}} = \dfrac{q}{t} = \dfrac{10}{2} = 5\) A. This equals the current at \(t = 1\) s, because \(i\) is linear in \(t\).

1. Charge is the area under the \(I\)-\(t\) graph. Triangle: \(\dfrac{1}{2} \times 2 \times 4 = 4\) C. Rectangle: \(3 \times 4 = 12\) C. Total \(16\) C.

2. \(I_{\text{av}} = \dfrac{16}{5} = 3.2\) A.

1. \(I = neAv_d\), so \(v_d = \dfrac{I}{neA} = \dfrac{1.7}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}} = \dfrac{1.7}{1.36 \times 10^{4}} = 1.25 \times 10^{-4}\) m/s.

2. \(t = \dfrac{1}{1.25 \times 10^{-4}} = 8000\) s, more than two hours.

Half the diameter means one quarter of the area. \(v_d = \dfrac{I}{neA}\) with the same \(I\) and \(n\), so \(v_d\) becomes \(4\) times larger. The same number of electrons per second must pass through a smaller area, so they must move faster.

1. Charge cannot pile up in the wire, so the same current flows through both parts: \(1 : 1\).

2. \(v_d = \dfrac{I}{neA}\). Part \(1\) has twice the area, so half the drift velocity: \(1 : 2\).

3. \(J = \dfrac{I}{A}\), so \(J_1 : J_2 = 1 : 2\).

4. \(E = \rho J\) with the same \(\rho\), so \(E_1 : E_2 = 1 : 2\). The field is stronger in the thinner part.

1. \(J = \dfrac{I}{A} = \dfrac{1.7}{10^{-6}} = 1.7 \times 10^{6}\) A/m\(^2\).

2. \(I = JA = 1.7 \times 10^{6} \times 2 \times 10^{-6} = 3.4\) A.

1. \(R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 100}{10^{-6}} = 1.7\ \Omega\).

2. \(I = \dfrac{V}{R} = \dfrac{3.4}{1.7} = 2\) A.

The volume \(AL\) stays the same. If \(L\) becomes \(kL\), then \(A\) becomes \(A/k\), and \(R = \dfrac{\rho L}{A}\) becomes \(k^2 R\).

1. \(k = 2\): \(R' = 4 \times 8 = 32\ \Omega\).

2. \(k = 1.5\): \(R' = 2.25 \times 8 = 18\ \Omega\).

1. On a \(V\)-\(I\) graph the slope is \(R\). \(R_A = \dfrac{8}{2} = 4\ \Omega\) and \(R_B = \dfrac{4}{4} = 1\ \Omega\).

2. \(B\) has the smaller resistance, so \(B\) carries more current for the same voltage. \(B\) is the better conductor.

\(R_T = R_0[1 + \alpha(T - T_0)] = 20[1 + 4 \times 10^{-3} \times 100] = 20 \times 1.4 = 28\ \Omega\).

\(R_T = R_0(1 + \alpha T)\), so \(5.6 = 5(1 + 30\alpha)\). Then \(30\alpha = 0.12\) and \(\alpha = 4 \times 10^{-3}\) per \(^\circ\)C.

1. \(J = \sigma E = 5 \times 10^{7} \times 0.1 = 5 \times 10^{6}\) A/m\(^2\).

2. \(I = JA = 5 \times 10^{6} \times 10^{-6} = 5\) A.

3. \(\rho = \dfrac{1}{\sigma} = 2 \times 10^{-8}\ \Omega\) m.

1. \(v_d = \dfrac{eE\tau}{m} = \dfrac{1.6 \times 10^{-19} \times 100 \times 2.25 \times 10^{-14}}{9 \times 10^{-31}} = \dfrac{3.6 \times 10^{-31}}{9 \times 10^{-31}} = 0.4\) m/s.

2. \(\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m} = \dfrac{0.4}{100} = 4 \times 10^{-3}\) m\(^2\)/(V s).

\(ne^2\tau = 8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 2.5 \times 10^{-14} = 8.5 \times 10^{28} \times 2.56 \times 10^{-38} \times 2.5 \times 10^{-14} = 5.44 \times 10^{-23}\).

\(\rho = \dfrac{9.1 \times 10^{-31}}{5.44 \times 10^{-23}} = 1.67 \times 10^{-8} \approx 1.7 \times 10^{-8}\ \Omega\) m, the measured value for copper.

1. The force on an electron is \(\vec F = -e\vec E\), opposite to the field, so towards \(P\). The electrons drift towards \(P\).

2. Conventional current is opposite to the electron drift, so from \(P\) to \(Q\), along \(\vec E\). The field points from high to low potential, so \(P\) is at the higher potential.

3. An electron accelerates only for a short time \(\tau\) before a collision with an ion. The collision resets its velocity in a random direction. The average of these short accelerations is the steady drift velocity.

In time \(t\) every electron drifts a distance \(v_d t\). So all the electrons in a cylinder of length \(v_d t\) and area \(A\) cross the end face in time \(t\).

Number of electrons in the cylinder \(= n \times A v_d t\). Charge \(= neAv_d t\).

\(I = \dfrac{\text{charge}}{t} = neAv_d\).

When the switch closes, the electric field is set up along the whole wire almost instantly, at nearly the speed of light. Free electrons everywhere in the circuit, including those inside the bulb, begin to drift at the same moment. The bulb does not wait for electrons from the switch to arrive.

\(\rho = \dfrac{m}{ne^2\tau}\). In a metal \(n\) is fixed. At a higher temperature the ions vibrate more, collisions are more frequent, \(\tau\) falls and \(\rho\) rises.

In a semiconductor heat frees many more charge carriers, so \(n\) rises steeply. This effect is far larger than the fall in \(\tau\), so \(\rho\) falls.

1. No. For Ohm's law the \(I\)-\(V\) graph must be a straight line through the origin. This graph bends.

2. \(R = \dfrac{V}{I}\). At \(P\): \(R = \dfrac{2}{1} = 2\ \Omega\). At \(Q\): \(R = \dfrac{6}{2} = 3\ \Omega\).

3. At \(Q\) the current is larger, so the filament is hotter. Resistance of a metal rises with temperature, so \(R\) is larger at \(Q\).

1. Current flows along the \(4\) cm length through the \(2 \times 1\) cm\(^2\) face. \(R = \dfrac{\rho L}{A} = \dfrac{4 \times 10^{-6} \times 0.04}{2 \times 10^{-4}} = 8 \times 10^{-4}\ \Omega\).

2. Now \(L = 1\) cm and \(A = 4 \times 2 = 8\) cm\(^2\). \(R = \dfrac{4 \times 10^{-6} \times 0.01}{8 \times 10^{-4}} = 5 \times 10^{-5}\ \Omega\).

3. Ratio \(= \dfrac{8 \times 10^{-4}}{5 \times 10^{-5}} = 16\). The long thin path resists far more than the short wide path.

\(P = VI\), so \(I = \dfrac{1000}{200} = 5\) A. \(R = \dfrac{V}{I} = \dfrac{200}{5} = 40\ \Omega\). Check: \(\dfrac{V^2}{R} = \dfrac{40000}{40} = 1000\) W.

1. \(P = I^2R = 4 \times 10 = 40\) W.

2. \(H = Pt = 40 \times 300 = 12000\) J \(= 12\) kJ.

1. \(R = \dfrac{V^2}{P} = \dfrac{200^2}{100} = 400\ \Omega\). \(I = \dfrac{P}{V} = \dfrac{100}{200} = 0.5\) A.

2. \(P = \dfrac{V^2}{R} = \dfrac{100^2}{400} = 25\) W. Half the voltage gives one quarter of the power.

1. \(1\) kWh \(= 1000\) W \(\times 3600\) s \(= 3.6 \times 10^{6}\) J.

2. Energy \(= 2\) kW \(\times 3\) h \(= 6\) kWh \(= 6\) units. Cost \(= 6 \times 5 =\) Rs \(30\).

1. At \(I = 2\) A the graph gives \(P = 20\) W. \(P = I^2R\), so \(R = \dfrac{20}{4} = 5\ \Omega\).

2. \(P = I^2R = 9 \times 5 = 45\) W.

3. \(P \propto I^2\), so the graph is a parabola. Doubling the current makes the power four times larger.

1. \(R = \dfrac{V^2}{P}\). \(R_A = \dfrac{100^2}{100} = 100\ \Omega\) and \(R_B = \dfrac{100^2}{50} = 200\ \Omega\).

2. Series resistance \(= 300\ \Omega\), so \(I = \dfrac{150}{300} = 0.5\) A. \(P_A = I^2R_A = 0.25 \times 100 = 25\) W and \(P_B = 0.25 \times 200 = 50\) W.

3. \(B\) produces its rated \(50\) W, so \(B\) glows normally. \(A\) produces only \(25\) W of its rated \(100\) W and glows dimly. In series the bulb with the smaller rating glows brighter.

In parallel each bulb has its rated \(100\) V, so each gives its rated power: \(100\) W and \(50\) W. Total \(= 150\) W. Total current \(= \dfrac{150}{100} = 1.5\) A.

1. \(R = \dfrac{V^2}{P}\) with the same \(V\), so \(R \propto \dfrac{1}{P}\). \(\dfrac{R_{60}}{R_{100}} = \dfrac{100}{60} = \dfrac{5}{3}\).

2. In series the current is the same, so \(P = I^2R \propto R\). The powers are in the ratio \(5 : 3\). The \(60\) W bulb has the larger resistance and glows brighter.

3. Total power \(= \dfrac{P_1P_2}{P_1 + P_2} = \dfrac{60 \times 100}{160} = 37.5\) W (see Question 20).

1. Each bulb gets \(100\) V, half its rated voltage. \(P = \dfrac{V^2}{R}\) becomes one quarter: \(25\) W each.

2. In parallel each bulb gets its full \(200\) V and gives \(100\) W. Total \(= 300\) W.

Let each coil have resistance \(R\). The heat needed is fixed: \(H = \dfrac{V^2}{R} \times 10\) min.

1. Series: resistance \(2R\), power halves, time doubles: \(20\) min.

2. Parallel: resistance \(\dfrac{R}{2}\), power doubles, time halves: \(5\) min.

Half the coil has half the resistance. \(P = \dfrac{V^2}{R}\) doubles. The same heat is produced in half the time.

1. \(R_2\) and \(R_3\) in parallel give \(2\ \Omega\). Total \(= 4 + 2 = 6\ \Omega\). \(I = \dfrac{12}{6} = 2\) A.

2. \(P_1 = I^2R_1 = 4 \times 4 = 16\) W. The \(2\) A splits equally: \(1\) A in each of \(R_2\) and \(R_3\), so \(P_2 = P_3 = 1 \times 4 = 4\) W. Total \(= 24\) W. Check: \(VI = 12 \times 2 = 24\) W.

3. \(R_1\) carries the full current and produces \(16\) W, four times as much as each of the others. \(R_1\) heats fastest.

\(H = I^2Rt = 16 \times 3 \times 420 = 20160\) J.

\(m = \dfrac{H}{L} = \dfrac{20160}{3.36 \times 10^{5}} = 0.06\) kg \(= 60\) g.

\(Q = mc\,\Delta T = 1 \times 4200 \times 50 = 210000\) J \(= 210\) kJ.

\(t = \dfrac{Q}{P} = \dfrac{210000}{500} = 420\) s \(= 7\) min.

Heat gained \(= 1 \times 4200 \times 70 = 294000\) J \(= 294\) kJ.

Energy supplied \(= 1000 \times 420 = 420000\) J \(= 420\) kJ.

Efficiency \(= \dfrac{294}{420} = 0.7 = 70\%\). The rest warms the vessel and the air.

1. \(I = \dfrac{P}{V}\). Heater: \(\dfrac{1000}{200} = 5\) A. Bulb: \(\dfrac{100}{200} = 0.5\) A.

2. Both on: the fuse carries \(5 + 0.5 = 5.5\) A, more than \(5\) A, so it melts. Heater alone: exactly \(5\) A, not more, so it does not melt.

1. \(P = I^2R\): \(1 = I^2 \times 100\), so \(I_{\max} = 0.1\) A. \(V_{\max} = IR = 0.1 \times 100 = 10\) V.

2. In series the same \(0.1\) A flows through both. Each can have \(10\) V, so the pair can have \(20\) V.

3. In parallel both have the same voltage, at most \(10\) V. Each then produces \(1\) W, so the total is \(2\) W.

1. Input \(= VI = 200 \times 5 = 1000\) W. Heat \(= 1000 - 800 = 200\) W.

2. Efficiency \(= \dfrac{800}{1000} = 80\%\). The heat is \(I^2R\): \(200 = 25R\), so \(R = 8\ \Omega\).

A charge \(q\) moving through a potential difference \(V\) has work \(W = qV\) done on it by the field. In a resistor this work becomes heat.

In time \(t\) the charge is \(q = It\) and \(V = IR\). So \(H = qV = It \times IR = I^2Rt\).

\(R_1 = \dfrac{V^2}{P_1}\) and \(R_2 = \dfrac{V^2}{P_2}\).

Series: \(R = R_1 + R_2 = V^2\left(\dfrac{1}{P_1} + \dfrac{1}{P_2}\right)\), so \(P = \dfrac{V^2}{R} = \dfrac{1}{1/P_1 + 1/P_2} = \dfrac{P_1P_2}{P_1 + P_2}\).

Parallel: each bulb has its rated \(V\), so \(P = P_1 + P_2\).

For \(40\) W and \(60\) W: series \(\dfrac{2400}{100} = 24\) W, parallel \(100\) W.

\(R \propto \dfrac{1}{r^2}\), so the thin wire has \(4\) times the resistance of the thick wire.

1. Parallel: same \(V\), \(P = \dfrac{V^2}{R} \propto \dfrac{1}{R}\). Thin to thick \(= 1 : 4\).

2. Series: same \(I\), \(P = I^2R \propto R\). Thin to thick \(= 4 : 1\).

1. At switching on: \(I = \dfrac{200}{40} = 5\) A. Glowing: \(I = \dfrac{200}{400} = 0.5\) A.

2. The filament is a metal, and its resistance rises with temperature. Glowing at about \(2500^\circ\)C, its resistance is ten times the cold value. The starting current is ten times the running current, and this surge breaks a weak filament.