The Workbook

1. Series: \(R = 2 + 3 = 5\ \Omega\).

2. Parallel: \(\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}\), so \(R = 1.2\ \Omega\). Less than the smaller one.

1. \(R = 3 \times 6 = 18\ \Omega\).

2. \(R = \dfrac{6}{3} = 2\ \Omega\).

3. Two in parallel give \(3\ \Omega\). In series with \(6\ \Omega\): \(R = 9\ \Omega\).

Series: \(R_s = nR\). Parallel: \(R_p = \dfrac{R}{n}\). Ratio \(\dfrac{R_s}{R_p} = n^2\). For \(n = 5\) the ratio is \(25\).

Both resistors are joined to the same two points, so the potential difference across each is \(V\).

The currents add at the junction: \(I = I_1 + I_2 = \dfrac{V}{R_1} + \dfrac{V}{R_2}\).

The combination has \(R = \dfrac{V}{I}\), so \(\dfrac{1}{R} = \dfrac{I}{V} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\).

1. \(\dfrac{1}{2} = \dfrac{1}{3} + \dfrac{1}{x}\), so \(\dfrac{1}{x} = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}\) and \(x = 6\ \Omega\).

2. \(3 + x = 5\), so \(x = 2\ \Omega\).

\(R_1 + R_2 = 9\) and \(\dfrac{R_1 R_2}{R_1 + R_2} = 2\), so \(R_1 R_2 = 18\).

Two numbers with sum \(9\) and product \(18\) are \(3\) and \(6\). So the resistors are \(3\ \Omega\) and \(6\ \Omega\).

In parallel the current divides in inverse ratio of the resistances: \(I_1 : I_2 = 12 : 4 = 3 : 1\).

\(I_1 = \dfrac{3}{4} \times 4 = 3\) A in the \(4\ \Omega\) and \(I_2 = 1\) A in the \(12\ \Omega\). Check: \(3 \times 4 = 1 \times 12 = 12\) V across both.

Each resistor has \(6\) V across it. \(I = \dfrac{6}{6} = 1\) A, \(\dfrac{6}{3} = 2\) A and \(\dfrac{6}{2} = 3\) A.

Total current \(= 1 + 2 + 3 = 6\) A. Equivalent resistance \(= \dfrac{6}{6} = 1\ \Omega\). Check: \(\dfrac{1}{6} + \dfrac{1}{3} + \dfrac{1}{2} = 1\).

1. \(6\ \Omega\) and \(3\ \Omega\) in parallel give \(\dfrac{6 \times 3}{9} = 2\ \Omega\). In series with \(4\ \Omega\): \(R = 6\ \Omega\).

2. \(I = \dfrac{18}{6} = 3\) A.

3. The parallel pair has \(3 \times 2 = 6\) V across it. Current in the \(6\ \Omega\): \(\dfrac{6}{6} = 1\) A. Current in the \(3\ \Omega\): \(\dfrac{6}{3} = 2\) A. Check: \(1 + 2 = 3\) A.

1. Balance condition: \(\dfrac{R_{AP}}{R_{PB}} = \dfrac{R_{AQ}}{R_{QB}}\). Here \(\dfrac{2}{4} = \dfrac{1}{2}\) and \(\dfrac{3}{6} = \dfrac{1}{2}\). Both arms divide the voltage in the same ratio, so \(V_P = V_Q\).

2. No potential difference across the \(5\ \Omega\), so its current is \(0\). It can be removed.

3. Upper arm: \(2 + 4 = 6\ \Omega\). Lower arm: \(3 + 6 = 9\ \Omega\). In parallel: \(R_{AB} = \dfrac{6 \times 9}{15} = 3.6\ \Omega\).

\(\dfrac{1}{R} = 1 + \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{11}{6}\), so \(R = \dfrac{6}{11}\ \Omega \approx 0.55\ \Omega\).

Largest: all in series, \(2 + 4 + 6 = 12\ \Omega\). Smallest: all in parallel, \(\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{11}{12}\), so \(R = \dfrac{12}{11}\ \Omega \approx 1.09\ \Omega\).

Each part has resistance \(\dfrac{9}{3} = 3\ \Omega\). Three \(3\ \Omega\) in parallel: \(\dfrac{3}{3} = 1\ \Omega\). In general, cutting into \(n\) parts and joining in parallel divides \(R\) by \(n^2\).

Volume is constant, so \(R \propto L^2\) and also \(R \propto \dfrac{1}{r^4}\).

1. Length doubles: \(R' = 4 \times 4 = 16\ \Omega\).

2. Radius halves: area becomes one quarter and length becomes \(4\) times. \(R' = 16 \times 4 = 64\ \Omega\).

Folding in half gives two wires of \(10\ \Omega\) each, side by side, in parallel: \(\dfrac{10}{2} = 5\ \Omega\).

Folding into four gives four wires of \(5\ \Omega\) in parallel: \(\dfrac{5}{4} = 1.25\ \Omega\). Folding into \(n\) divides \(R\) by \(n^2\), like cutting.

1. \(R = \dfrac{\rho L}{\pi r^2}\), so \(R \propto \dfrac{1}{r^2}\). Thin to thick: \(\dfrac{(2r)^2}{r^2} = 4 : 1\).

2. The thick wire has \(1\ \Omega\). In parallel: \(\dfrac{4 \times 1}{5} = 0.8\ \Omega\).

1. The ring splits into two arcs in parallel: a quarter, \(3\ \Omega\), and three quarters, \(9\ \Omega\). \(R = \dfrac{3 \times 9}{12} = 2.25\ \Omega\).

2. Two half rings of \(6\ \Omega\) each in parallel: \(R = 3\ \Omega\).

The two half rings have \(\dfrac{R}{2}\) each and are in parallel: \(\dfrac{R/2 \times R/2}{R} = \dfrac{R}{4}\).

For points dividing the ring \(1 : 2\), the arcs are \(\dfrac{R}{3}\) and \(\dfrac{2R}{3}\). In parallel: \(\dfrac{(R/3)(2R/3)}{R} = \dfrac{2R}{9}\).

1. Between \(A\) and \(C\) there are three paths in parallel. The path \(A\)-\(B\)-\(C\) is \(12\ \Omega\), the path \(A\)-\(D\)-\(C\) is \(12\ \Omega\) and the diagonal is \(6\ \Omega\). \(\dfrac{1}{R} = \dfrac{1}{12} + \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1}{3}\), so \(R = 3\ \Omega\).

2. Between \(A\) and \(B\): the direct \(6\ \Omega\) is in parallel with the route \(A\) to \(C\) to \(B\). From \(A\) to \(C\) the diagonal (\(6\)) is in parallel with \(A\)-\(D\)-\(C\) (\(12\)): \(4\ \Omega\). Then \(C\) to \(B\) adds \(6\): the route is \(10\ \Omega\). \(R = \dfrac{6 \times 10}{16} = 3.75\ \Omega\).

The bridge is balanced because \(\dfrac{R}{R} = \dfrac{R}{R}\). The two middle points are at the same potential, so no current flows in the middle resistor and it can be removed.

Each arm is then \(2R\), and two \(2R\) in parallel give \(R\).

1. The ladder is infinite. Removing one section from the front leaves a ladder with exactly the same pattern, so its resistance is still \(x\).

2. The first \(1\ \Omega\) is in series with the first \(2\ \Omega\) rung in parallel with \(x\): \(x = 1 + \dfrac{2x}{2 + x}\).

   \(x(2 + x) = 2 + x + 2x\), so \(x^2 - x - 2 = 0\) and \((x - 2)(x + 1) = 0\). The positive root gives \(x = 2\ \Omega\).

Let the input resistance be \(x\). Removing the first section leaves \(x\) again, so \(x = R + \dfrac{Rx}{R + x}\).

\(x(R + x) = R(R + x) + Rx\), so \(x^2 - Rx - R^2 = 0\) and \(x = \dfrac{R + \sqrt{R^2 + 4R^2}}{2} = \dfrac{(1 + \sqrt 5)}{2}R \approx 1.62R\).

1. The three edges at \(A\) look alike, so the current splits equally: \(\dfrac{I}{3}\) in each. At \(B\) the current \(\dfrac{I}{3}\) splits equally into two middle edges: \(\dfrac{I}{6}\) each. The three edges at \(G\) carry \(\dfrac{I}{3}\) each again.

2. Along \(A\)-\(B\)-\(C\)-\(G\): \(V = \dfrac{I}{3}r + \dfrac{I}{6}r + \dfrac{I}{3}r = \dfrac{5Ir}{6}\).

3. \(R = \dfrac{V}{I} = \dfrac{5r}{6}\).

Current in \(= 3 + 5 = 8\) A. Current out so far \(= 2\) A. The fourth wire must carry \(6\) A out of the junction.

1. In \(= 4\) A. Out \(= 1 + 2 + I\). So \(I = 4 - 3 = 1\) A, leaving as drawn.

2. A negative value means the arrow was drawn the wrong way. The current has the size found, but it enters the junction.

The total current entering a junction equals the total current leaving it.

It follows from conservation of charge. A junction cannot create or destroy charge, and in a steady current it cannot store charge either.

If more current entered than left, charge would pile up at the junction. Its potential would rise without limit, which does not happen in a steady state.

The algebraic sum of the potential changes around any closed loop is zero.

It follows from conservation of energy. Potential is the energy per unit charge, and a charge carried around a loop returns to the same point with the same energy.

1. \(-IR\). Moving with the current means moving from high to low potential.

2. \(+IR\).

3. \(+E\). The positive terminal is higher by \(E\) (ideal cell).

4. \(-E\).

1. Take the current \(I\) clockwise, out of the positive plate of the \(12\) V cell. Going clockwise: \(+12 - 2I - 3 - 4I = 0\), so \(6I = 9\) and \(I = 1.5\) A.

2. \(V = IR = 1.5 \times 4 = 6\) V.

3. The current enters the \(3\) V cell at its positive plate, so that cell is being charged.

Now the emfs add: \(12 + 3 - 2I - 4I = 0\), so \(I = \dfrac{15}{6} = 2.5\) A. Across the \(4\ \Omega\): \(2.5 \times 4 = 10\) V.

1. Walk from \(A\) to \(B\). Across the resistor, with the current: \(-2 \times 3 = -6\) V. Across the cell from \(+\) to \(-\): \(-6\) V. So \(V_B = V_A - 12\), and \(V_A - V_B = 12\) V.

2. With the cell reversed we cross it from \(-\) to \(+\): \(+6\) V. \(V_A - V_B = 6 - 6 = 0\).

\(V_B = V_A - IR + E = 10 - 2 \times 4 + 5 = 7\) V.

1. Left branch, from the \(12\) V cell up to \(N\): \(I_1 = \dfrac{12 - V}{1}\). Right branch: \(I_2 = \dfrac{6 - V}{1}\). Middle branch, down from \(N\) to \(O\): \(I_3 = \dfrac{V}{4}\).

2. At \(N\): \(I_1 + I_2 = I_3\). So \((12 - V) + (6 - V) = \dfrac{V}{4}\). Then \(72 - 8V = V\) and \(V = 8\) V.

3. \(I_1 = 4\) A, \(I_3 = 2\) A, \(I_2 = 6 - 8 = -2\) A. The minus sign means \(2\) A flows down through the \(6\) V branch, into the cell's positive plate.

1. The \(6\) V cell. The current enters it at the positive plate, so the cell absorbs energy.

2. Delivered: \(12 \times 4 = 48\) W. Absorbed by the \(6\) V cell: \(6 \times 2 = 12\) W. Heat: \(4^2 \times 1 + 2^2 \times 4 + 2^2 \times 1 = 16 + 16 + 4 = 36\) W. Check: \(48 = 12 + 36\).

\((12 - V) + (6 - V) = \dfrac{V}{2}\), so \(36 - 4V = V\) and \(V = 7.2\) V.

\(I_1 = 12 - 7.2 = 4.8\) A, \(I_3 = \dfrac{7.2}{2} = 3.6\) A, \(I_2 = 6 - 7.2 = -1.2\) A. Check: \(4.8 - 1.2 = 3.6\).

Step 1: choose one node as the reference and call its potential zero. Step 2: give each other node a potential symbol. Write the current in every branch as (potential difference)/(resistance), including the emf of any cell in the branch. Step 3: apply the junction rule at each unknown node and solve.

Parallel branches share the same two nodes, so one unknown handles all of them. The loop method needs one unknown current per branch and one equation per loop.

1. Currents towards \(N\): \(\dfrac{10 - V}{1}\) and \(\dfrac{4 - V}{2}\). Current away through the load: \(\dfrac{V}{2}\). So \(10 - V + \dfrac{4 - V}{2} = \dfrac{V}{2}\). Multiply by \(2\): \(20 - 2V + 4 - V = V\), so \(V = 6\) V.

2. \(10\) V branch: \(10 - 6 = 4\) A upward. \(4\) V branch: \(\dfrac{4 - 6}{2} = -1\) A, so \(1\) A flows downward into the positive plate of the \(4\) V cell. Load: \(\dfrac{6}{2} = 3\) A. Check: \(4 - 1 = 3\).

Delivered: \(10 \times 4 = 40\) W. Absorbed by the \(4\) V cell: \(4 \times 1 = 4\) W.

Heat: \(4^2 \times 1 + 1^2 \times 2 + 3^2 \times 2 = 16 + 2 + 18 = 36\) W. Check: \(40 = 4 + 36\).

Let the common voltage be \(V\). Currents towards the top node: \(\dfrac{2 - V}{1}\) and \(\dfrac{4 - V}{1}\); through the load: \(\dfrac{V}{2}\).

\(2 - V + 4 - V = \dfrac{V}{2}\), so \(12 - 4V = V\) and \(V = 2.4\) V.

\(2\) V cell: \(2 - 2.4 = -0.4\) A, so \(0.4\) A flows into the cell (it is being charged). \(4\) V cell: \(1.6\) A out. Load: \(1.2\) A. Check: \(1.6 - 0.4 = 1.2\).

Going around the loop the two emfs cancel: \(E - E - IR = 0\), so \(I = 0\). No current means no potential difference across the resistor.

Let the voltage across the pair be \(V\), with nothing connected outside. Current from cell 1 towards the positive node: \(\dfrac{E_1 - V}{r_1}\). From cell 2: \(\dfrac{E_2 - V}{r_2}\).

No current leaves the pair, so the two currents add to zero: \(\dfrac{E_1 - V}{r_1} + \dfrac{E_2 - V}{r_2} = 0\).

\(\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2} = V\left(\dfrac{1}{r_1} + \dfrac{1}{r_2}\right)\), so \(V = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}\).

1. Total resistance \(= 1 + 2 + 3 = 6\ \Omega\). \(I = \dfrac{12}{6} = 2\) A.

2. Start at \(R\) (\(0\) V) and go backwards against the current to the positive plate: \(V_S = 12\) V. Then move with the current: \(V_P = 12 - 2 \times 1 = 10\) V, \(V_Q = 10 - 2 \times 2 = 6\) V, \(V_R = 6 - 2 \times 3 = 0\) V. Consistent.

Earthing only fixes the zero; the differences stay the same. With \(V_Q = 0\): \(V_P = 4\) V, \(V_S = 6\) V, \(V_R = -6\) V.

The current does not change. The earth connection carries no current, and the loop equation does not depend on where the zero is.

From \(S\) with the current: \(-2 \times 1 - 2 \times 2 - 2 \times 3 = -12\) V across the three resistors. Then through the cell from \(-\) to \(+\): \(+12\) V.

Sum \(= -2 - 4 - 6 + 12 = 0\), as the loop rule requires.

1. \(\tau = RC = 2 \times 10^{3} \times 500 \times 10^{-6} = 1\) s.

2. \(\Omega \times\) F \(= \dfrac{\text{V}}{\text{A}} \times \dfrac{\text{C}}{\text{V}} = \dfrac{\text{C}}{\text{A}} = \dfrac{\text{C}}{\text{C/s}} =\) s.

1. \(q_{\max} = CE = 500 \times 10^{-6} \times 12 = 6 \times 10^{-3}\) C \(= 6\) mC.

2. At \(t = \tau = 1\) s: \(q = 6(1 - 0.37) = 6 \times 0.63 = 3.78 \approx 3.8\) mC.

\(i_0 = \dfrac{E}{R} = \dfrac{12}{2000} = 6\) mA. At \(t = 1\) s \(= \tau\): \(i = 6 \times 0.37 = 2.2\) mA. After a long time the capacitor is full and \(i = 0\).

1. \(1 - e^{-t/\tau} = \dfrac{1}{2}\), so \(e^{-t/\tau} = \dfrac{1}{2}\) and \(t = \tau \ln 2 = 0.69\) s.

2. \(e^{-t/\tau} = 0.01\), so \(t = \tau \ln 100 = 4.6\) s. After about five time constants the capacitor is practically full.

1. The curve levels off at \(12\) mC. \(C = \dfrac{q_{\max}}{E} = \dfrac{12 \times 10^{-3}}{6} = 2 \times 10^{-3}\) F \(= 2\) mF.

2. At \(t = \tau\) the charge is \((1 - 1/e) = 63\%\) of the maximum. So \(\tau = 2\) s.

3. \(R = \dfrac{\tau}{C} = \dfrac{2}{2 \times 10^{-3}} = 1000\ \Omega = 1\) k\(\Omega\).

1. \(\tau = 10^{5} \times 10^{-5} = 1\) s. \(q_0 = CV = 10^{-5} \times 100 = 10^{-3}\) C \(= 1\) mC. \(i_0 = \dfrac{V}{R} = \dfrac{100}{10^{5}} = 1\) mA.

2. \(q = q_0 e^{-t/\tau} = 1 \times e^{-2} = 0.135\) mC.

1. At \(t = 0\) the graph gives \(i_0 = 4\) mA. The current is \(i_0/e\) at \(t = \tau\), so \(\tau = 3\) s.

2. \(R = \dfrac{\tau}{C} = \dfrac{3}{300 \times 10^{-6}} = 10^{4}\ \Omega = 10\) k\(\Omega\). \(V_0 = i_0 R = 4 \times 10^{-3} \times 10^{4} = 40\) V.

1. \(e^{-t/\tau} = \dfrac{1}{2}\) gives \(t = \tau \ln 2 = 0.69\) s.

2. One quarter is two halvings: \(t = 2 \times 0.69 = 1.38\) s. Each halving takes the same time, like radioactive decay.

The charge falls to \(\dfrac{1}{e}\) in one time constant, so \(\tau = 5\) s. \(R = \dfrac{\tau}{C} = \dfrac{5}{10^{-4}} = 5 \times 10^{4}\ \Omega = 50\) k\(\Omega\).

1. At \(t = 0\) the capacitor has no charge and no voltage. It acts as a wire across \(R_2\), so \(R_2\) carries no current. \(I = \dfrac{12}{2} = 6\) A.

2. After a long time no current flows into the capacitor. \(R_1\) and \(R_2\) are in series: \(I = \dfrac{12}{6} = 2\) A.

3. The capacitor is across \(R_2\): \(V_C = 2 \times 4 = 8\) V. \(q = CV = 3 \times 8 = 24\ \mu\)C.

At \(t = 0\) the capacitor is uncharged, so the potential difference across it is zero, exactly like a wire. Current flows into it freely.

At \(t = \infty\) the capacitor is full. Its voltage balances the rest of the circuit and no more current flows into it, exactly like a gap.

Same in both cases: no current can pass between the plates. Different: at \(t = 0\) the voltage is zero, at \(t = \infty\) the current is zero.

1. \(V_C = 0\) (uncharged), so \(V_R = E = 10\) V.

2. \(V_C = E(1 - e^{-1}) = 10 \times 0.63 = 6.3\) V. \(V_R = E e^{-1} = 3.7\) V.

3. \(V_C = 10\) V and \(V_R = 0\). By the loop rule \(V_R + V_C = E = 10\) V at every instant.

1. \(\tau = 10^{4} \times 10^{-4} = 1\) s. \(q_{\max} = CE = 10^{-4} \times 10 = 10^{-3}\) C \(= 1000\ \mu\)C. At \(t = \tau\): \(q = 0.63 \times 1000 = 630\ \mu\)C and \(i = \dfrac{10}{10^{4}} \times 0.37 = 0.37\) mA.

2. \(900 = 1000(1 - e^{-t})\), so \(e^{-t} = 0.1\) and \(t = \ln 10 = 2.3\) s.

1. The battery moves charge \(q = CE\) through \(E\): \(W = CE^2 = 500 \times 10^{-6} \times 144 = 0.072\) J \(= 72\) mJ.

2. \(U = \dfrac{1}{2}CE^2 = 36\) mJ.

3. Heat \(= 72 - 36 = 36\) mJ, exactly half. This is true for any \(R\); the resistance only sets how fast the heat is produced.

1. \(R = 400 + 600 = 1000\ \Omega\). \(\tau = 1000 \times 10^{-3} = 1\) s. \(i_0 = \dfrac{100}{1000} = 0.1\) A.

2. At \(t = 0\): \(V = 0.1 \times 600 = 60\) V. At \(t = 1\) s \(= \tau\): \(i = 0.037\) A, so \(V = 0.037 \times 600 = 22\) V.

All the stored energy becomes heat: \(H = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 10^{-3} \times 10^{4} = 5\) J.

The same current flows through both resistors at every instant, so the heat divides in the ratio of the resistances, \(400 : 600 = 2 : 3\). Heat in \(400\ \Omega\): \(2\) J. Heat in \(600\ \Omega\): \(3\) J.

At \(t = 0\) the capacitor has no voltage, so the whole emf is across \(R\) and \(i_0 = \dfrac{E}{R}\). The capacitor plays no part yet.

At \(t = \infty\) the current is zero, so there is no voltage across \(R\). The whole emf is across the capacitor and \(q = CE\). The resistance only decides how long this takes.

\(\tau = 10^{6} \times 2 \times 10^{-6} = 2\) s. With \(2R\) and \(\dfrac{C}{2}\) the product \(RC\) is unchanged, so \(\tau\) stays \(2\) s.

\(\dfrac{dq}{dt} = CE \times \dfrac{1}{RC}e^{-t/RC} = \dfrac{E}{R}e^{-t/RC}\).

Substitute: \(E - \dfrac{q}{C} - R\dfrac{dq}{dt} = E - E(1 - e^{-t/RC}) - E e^{-t/RC} = E - E + E e^{-t/RC} - E e^{-t/RC} = 0\).

\(RC = \dfrac{V}{I} \times \dfrac{q}{V} = \dfrac{q}{I}\). Current is charge per time, so \(\dfrac{q}{I}\) is a time. \([RC] = \dfrac{[\text{AT}]}{[\text{A}]} = [\text{T}]\).

With the battery replaced by a wire, the capacitor sees \(R_2\) in parallel with \(R_1\): \(R = \dfrac{2 \times 4}{6} = \dfrac{4}{3}\ \Omega\).

\(\tau = RC = \dfrac{4}{3} \times 3 \times 10^{-6} = 4\ \mu\)s.