Electrical circuits

Ohm's law alone cannot solve a circuit with two cells in different branches, or a bridge, or a network that is neither series nor parallel. This unit gives you the two rules that solve any network — Kirchhoff's laws — and then applies them to the three instruments the NEB paper asks about every year: the Wheatstone bridge, the metre bridge and the potentiometer. It ends with what current actually does in a resistor: it heats it, according to Joule's law.

14.1 Kirchhoff's law

A junction (or node) is a point where three or more conductors meet. A loop is any closed conducting path. Kirchhoff gave one rule for each.

Sign rules that never fail. Choose a direction to walk round the loop, then:

Element met while walking round	Term to write
Resistor traversed along the assumed current	-IR
Resistor traversed against the assumed current	+IR
Cell entered at - and left at +	+E
Cell entered at + and left at -	-E

Set the algebraic sum to zero. If a current comes out negative, its true direction is opposite to the one you assumed — the magnitude is still right, so never go back and redraw.

14.2 Wheatstone bridge circuit; Meter bridge

Four resistors P, Q, R, S form a quadrilateral ABCD. A cell is across one diagonal AC and a galvanometer across the other, BD. The bridge is balanced when the galvanometer shows no deflection.

The metre bridge

The metre bridge is the Wheatstone bridge built from a single uniform wire. A manganin or constantan wire exactly 100\ \text{cm} long is stretched along a metre scale between copper strips. The unknown resistance X goes in the left gap and a resistance box R in the right gap. A jockey slides along the wire until the galvanometer reads zero at a balance length l from the left end.

If the wire has resistance \sigma per centimetre, the two parts of the wire have resistances \sigma l and \sigma(100-l). Applying the balance condition,

14.3 Potentiometer: Comparison of e.m.f.; measurement of internal resistance of a cell

A potentiometer is a long uniform wire (usually 4 m or 10 m, wound in 1 m strips) carrying a steady current from a driver cell of e.m.f. E_0. The potential falls uniformly along it, so the potential difference across a length l is V = kl, where the potential gradient

is constant as long as the driver current is steady.

Comparison of e.m.f. of two cells

Connect the two cells E_1 and E_2 through a two-way key so that either can be put in the galvanometer branch, with their positive terminals joined to the same end A as the driver cell. If they balance at lengths l_1 and l_2,

Internal resistance of a cell

14.4 Super conductors; Perfect conductors

The resistivity of a normal metal falls as it is cooled, because the lattice ions vibrate less and scatter electrons less. But it never reaches zero: impurities and defects leave a residual resistivity even at 0\ \text{K}.

Certain materials behave completely differently. Below a sharp critical temperature T_c their resistivity drops abruptly to zero — not "very small", but zero, as far as any measurement can tell. Kamerlingh Onnes discovered this in mercury in 1911 at T_c = 4.2\ \text{K}.

Perfect conductor versus superconductor. A perfect conductor is an idealisation: a material with \rho = 0 but with no other special property. The distinction is not academic, and examiners like it:

	Perfect conductor (hypothetical)	Superconductor (real)
Resistivity	zero	zero
Defined by	\rho = 0 only	\rho = 0 and B = 0 inside
Cooled in a magnetic field	traps the flux already inside it, because d\vec{B}/dt = 0	expels the flux completely (Meissner effect)
Final state	depends on the history of cooling and magnetising	same state whatever the order — a true thermodynamic phase
Exists?	no	yes, below T_c

A superconductor is destroyed by too large a magnetic field (the critical field B_c, which falls to zero at T_c) or by too large a current (the critical current). Both limits matter in real magnets.

High-temperature superconductors. In 1986–87 ceramic oxides such as YBa₂Cu₃O₇ were found to superconduct up to about 92 K — above the boiling point of liquid nitrogen (77 K), which is cheap. Uses: MRI and NMR magnets, particle-accelerator magnets, maglev trains, SQUID magnetometers, and loss-free power cables.

14.5 Conversion of galvanometer into voltmeter and ammeter; Ohmmeter

A moving-coil galvanometer has a resistance G and gives full-scale deflection for a small current I_g (typically a few milliamperes). It is the movement inside every analogue meter; only the resistor bolted to it changes.

The ohmmeter

An ohmmeter is a galvanometer, a cell of e.m.f. E and an adjustable resistor R in series, with two test leads. First the leads are touched together (zero external resistance) and R is adjusted so that the deflection is exactly full scale:

With an unknown resistance X between the leads the current becomes

so each deflection corresponds to one value of X. Two features follow at once and are often asked:

• the scale is non-uniform (cramped at large X), because I is not proportional to X;
• the scale is reversed: zero ohms is at the full-scale end (right) and infinite ohms at the zero-current end (left).

14.6 Joule's law

When a current passes through a conductor, the ordered drift energy the electrons gain from the field is handed to the lattice in collisions, and the conductor warms up.

The three statements usually quoted as Joule's laws are just the three proportionalities hidden in H = I^2Rt:

1. H \propto I^{2} when R and t are constant;
2. H \propto R when I and t are constant;
3. H \propto t when I and R are constant.

If the heat is wanted in calories, divide by the mechanical equivalent of heat J = 4.2\ \text{J cal}^{-1}: H = I^2Rt/J calories.

Where it is used. Heaters, immersion rods, electric irons and incandescent lamps use a high-resistivity alloy (nichrome, \rho \approx 1.1\times10^{-6}\ \Omega\,\text{m}) with a high melting point. A fuse does the opposite job: a short piece of low-melting tin–lead alloy that melts and breaks the circuit when I^2R heating becomes dangerous. Electrical energy is sold in kilowatt-hours: 1\ \text{kWh} = 1000 \times 3600 = 3.6\times10^{6}\ \text{J}.

Chapter summary

• Kirchhoff's junction law \sum I = 0 expresses conservation of charge; the loop law \sum E = \sum IR expresses conservation of energy.
• A Wheatstone bridge is balanced when I_g = 0, and then P/Q = R/S. The condition is independent of the cell and of the galvanometer.
• Metre bridge: X = R\,l/(100-l), with l measured from the end next to X. Keep l near 50 cm and interchange X and R to cancel end errors.
• Potentiometer: V = kl with k = V_{AB}/L. Comparison of e.m.f. E_1/E_2 = l_1/l_2; internal resistance r = R(l_1-l_2)/l_2. At balance the cell supplies no current, so the true e.m.f. is measured.
• A superconductor has \rho = 0 below T_c and expels magnetic flux (Meissner effect); a merely perfect conductor would trap the flux instead.
• Galvanometer conversions: ammeter S = I_gG/(I-I_g) in parallel; voltmeter R = V/I_g - G in series. An ohmmeter has a reversed, non-uniform scale.
• Joule's law: H = I^2Rt = VIt = V^2t/R; power P = I^2R = V^2/R; 1\ \text{kWh} = 3.6\times10^{6}\ \text{J}.

Practice questions

Group A — Multiple choice (1 mark each)

1. Kirchhoff's junction law is a consequence of the conservation of [1]aenergybchargecmomentumdmagnetic flux
2. In a balanced Wheatstone bridge, the current through the galvanometer is [1]amaximumbzerochalf the main currentdequal to the main current
3. A potentiometer measures the e.m.f. of a cell more accurately than a voltmeter because at balance it [1]ahas a very low resistancebdraws no current from the cellcuses a longer wiredis calibrated in volts
4. To convert a galvanometer into an ammeter, we connect a [1]ahigh resistance in seriesbhigh resistance in parallelclow resistance in seriesdlow resistance in parallel
5. Two resistors of 2\ \Omega and 4\ \Omega are joined in parallel across a battery. The ratio of the heat produced in them in a given time is [1]a1:2b2:1c1:4d4:1
6. Below its critical temperature a superconductor [1]ahas infinite resistancebbecomes an insulatorcexpels magnetic flux from its interiordtraps all the flux inside it

Answers to Group A

1. (b) — charge cannot accumulate at a junction, so what flows in must flow out.

2. (b) — balance is defined by the null (zero) galvanometer deflection.

3. (b) — with no current drawn there is no internal drop Ir, so the reading is the true e.m.f.

4. (d) — the shunt S = I_gG/(I-I_g) is small and goes in parallel.

5. (b) — in parallel V is common, so H = V^2t/R \propto 1/R, giving H_2:H_4 = \frac14 : \frac12 inverted, i.e. 2:1.

6. (c) — the Meissner effect; that, plus \rho = 0, defines superconductivity.

Group B — Short answer (5 marks each)

1. State Kirchhoff's laws and name the conservation principle behind each. Using them, obtain the balance condition of a Wheatstone bridge. [5]
2. In a metre bridge the unknown resistance X is in the left gap and a 6\ \Omega resistance in the right gap. The balance point is 40\ \text{cm} from the left end. Find X and the new balance length if the two are interchanged. [5]
3. A cell balances against 150\ \text{cm} of a potentiometer wire. When a 4\ \Omega resistor is connected across the cell, the balance length becomes 120\ \text{cm}. Calculate the internal resistance of the cell. [5]
4. A galvanometer of resistance 100\ \Omega gives full-scale deflection for 1\ \text{mA}. Find the resistance needed to convert it into(a) an ammeter reading up to 1\ \text{A} and(b) a voltmeter reading up to 5\ \text{V}. [5]
5. Distinguish between a perfect conductor and a superconductor. Why is a superconducting magnet cooled below T_c before the field is switched on? [5]

Answers to Group B

1. Junction law \sum I = 0 (conservation of charge); loop law \sum E = \sum IR (conservation of energy). At balance I_g = 0, so I_1 flows through P and Q and I_2 through R and S, and V_B = V_D. Then I_1P = I_2R and I_1Q = I_2S; dividing gives P/Q = R/S.

2. X = R\,l/(100-l) = 6 \times 40/60 = 4\ \Omega. Interchanged, the left gap holds 6\ \Omega, so 6/4 = l'/(100-l'), giving 600 - 6l' = 4l' and l' = 60\ \text{cm}.

3. r = R\left(\dfrac{l_1-l_2}{l_2}\right) = 4\left(\dfrac{150-120}{120}\right) = 4 \times 0.25 = 1\ \Omega.

4. (a) S = \dfrac{I_gG}{I-I_g} = \dfrac{(10^{-3})(100)}{1-0.001} = \dfrac{0.1}{0.999} = 0.100\ \Omega in parallel. (b) R = \dfrac{V}{I_g} - G = \dfrac{5}{10^{-3}} - 100 = 5000 - 100 = 4900\ \Omega in series.

5. A perfect conductor only has \rho = 0; a superconductor also has B = 0 inside it. If a perfect conductor were cooled in a field it would trap that flux, because d\vec{B}/dt = 0 inside a resistanceless material; a superconductor expels the flux (Meissner effect) whatever the order of cooling and magnetising. In practice the magnet is cooled first so that the whole winding is superconducting before any current — and hence any I^2R heating at a normal spot — can appear; a normal region would heat, spread, and "quench" the magnet.

Group C — Long answer (8 marks each)

1. (a) Draw a labelled diagram of a potentiometer arranged to compare the e.m.f.s of two cells, and derive E_1/E_2 = l_1/l_2. [4](b) A Daniell cell of e.m.f. 1.08\ \text{V} balances at 216\ \text{cm}. A second cell balances at 300\ \text{cm} on the same wire. Find the e.m.f. of the second cell and the potential gradient of the wire. [4]
2. (a) State and derive Joule's law of heating, and explain why the element of a heater glows red while its copper connecting leads stay cool. [4](b) An electric heater marked "1000 W, 220 V" is used for 5 hours a day for 30 days. Find its resistance, the current it draws, and the monthly cost of running it at NPR 10 per unit. [4]

Answers to Group C

1. (b) Since E \propto l for the same wire and driver current,

E_2 = E_1\frac{l_2}{l_1} = 1.08 \times \frac{300}{216} = 1.50\ \text{V}

Potential gradient k = E_1/l_1 = 1.08/216 = 5.0\times10^{-3}\ \text{V cm}^{-1} = 0.50\ \text{V m}^{-1}.

2. (a) W = qV = VIt = I^2Rt, all of which appears as heat since the drift velocity is unchanged. In a series path the current is the same everywhere, so H \propto R: the nichrome element has a resistance of tens of ohms and a high melting point, while the thick copper leads have a resistance of a fraction of an ohm, so almost all the heat is produced in the element.

(b) R = V^2/P = 220^2/1000 = 48.4\ \Omega; I = P/V = 1000/220 = 4.55\ \text{A}. Energy = 1.0\ \text{kW} \times 5\ \text{h} \times 30 = 150\ \text{kWh}, so the cost is 150 \times 10 = NPR 1500 per month.