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CalibratePhysics

Chapter 9

Current Electricity

Maharashtra HSC · 5 subtopics · textbook pages 214–229

Formula SheetEvery formula this chapter can be examined on, in one table.

Symbols, units and conventions

  • II current in ampere (A), VV potential difference in volt (V), EE (written ε\varepsilon in the textbook) emf in volt, RR, rr, GG, SS, XX resistances in ohm (Ω\Omega), ρ\rho specific resistance (resistivity) in Ω m\Omega\,\text{m}, KK potential gradient in V m−1\text{V m}^{-1}.
  • Junction: a point where two or more conductors meet. Loop (mesh): any closed conducting path. Branch: any part of a network lying between two junctions.
  • Sign convention for the current law: currents arriving at a junction are positive, currents leaving it are negative.
  • Sign convention for the voltage law: tracing a resistor along the direction of the conventional current gives −IR-IR; tracing it against the current gives +IR+IR. Tracing a source from its negative to its positive terminal gives +E+E; from positive to negative gives −E-E.
  • Useful conversions: 1 mA=10−3 A1\ \text{mA} = 10^{-3}\ \text{A}, 1 μA=10−6 A1\ \mu\text{A} = 10^{-6}\ \text{A}, 1 cm=10−2 m1\ \text{cm} = 10^{-2}\ \text{m}. A metre-bridge wire is exactly 100100 cm long.

Recall from Std XI (needed for almost every numerical)

FormulaSymbols, units and conditions
V=IRR=ρLA=ρLπr2V = IR \qquad R = \dfrac{\rho L}{A} = \dfrac{\rho L}{\pi r^2}
  • Ohm's law, and the resistance of a uniform wire of length LL, area of cross section AA and radius rr. Rearranged, ρ=R πr2L\rho = \dfrac{R\,\pi r^2}{L} — this is how the specific resistance of a bridge wire is found.
Rs=R1+R2+R3+…1Rp=1R1+1R2+1R3+…R_s = R_1 + R_2 + R_3 + \dots \qquad \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots
  • Series and parallel combinations. For just two resistors in parallel, Rp=R1R2R1+R2R_p = \dfrac{R_1R_2}{R_1+R_2}.
I=ER+rV=E−IrI = \dfrac{E}{R+r} \qquad V = E - Ir
  • Current drawn from a cell of emf EE and internal resistance rr through an external resistance RR, and the terminal potential difference VV across the cell while it supplies that current. V<EV<E whenever current flows; V=EV=E only on open circuit.

Kirchhoff's laws of electrical network

FormulaSymbols, units and conditions
∑i=1nIi=0\sum_{i=1}^{n} I_i = 0
  • Kirchhoff's first law (current law, junction law): the algebraic sum of the currents at a junction in an electrical network is zero, where IiI_i is the current in the ithi^{th} conductor at a junction having nn conductors. Equivalently, total current arriving equals total current leaving. It expresses conservation of electric charge.
∑IR+∑E=0\sum IR + \sum E = 0
  • Kirchhoff's second law (voltage law, loop theorem): the algebraic sum of the potential differences (products of current and resistance) and the emfs in a closed loop is zero. It expresses conservation of energy. Applies to any loop of a network, however complicated, and to both DC and AC circuits.

Wheatstone bridge

FormulaSymbols, units and conditions
PQ=SRequivalentlyPR=QS\dfrac{P}{Q} = \dfrac{S}{R} \qquad \text{equivalently} \qquad PR = QS
  • Balancing condition for a Wheatstone network ABCD with PP in AB, QQ in BC, RR in CD and SS in DA, the galvanometer between B and D and the cell between A and C. "Balanced" means the galvanometer current Ig=0I_g = 0. Knowing any three resistances, the fourth follows. Used for resistances from about tens of ohm to hundreds of ohm.
I=I1+I2−I1P−IgG+I2S=0−(I1−Ig)Q+(I2+Ig)R+IgG=0I = I_1 + I_2 \qquad -I_1P - I_gG + I_2S = 0 \qquad -(I_1-I_g)Q + (I_2+I_g)R + I_gG = 0
  • The three equations of the unbalanced bridge (junction A, loop ABDA clockwise, loop BCDB clockwise). Solving them together gives the current in any branch.

Metre bridge

FormulaSymbols, units and conditions
XR=RADRDB=lxlR∴ X=lxlRR\dfrac{X}{R} = \dfrac{R_{AD}}{R_{DB}} = \dfrac{l_x}{l_R} \qquad \therefore\ X = \dfrac{l_x}{l_R}R
  • Unknown resistance XX in the left gap, resistance box RR in the right gap, null point D on the wire; lxl_x is the length AD (left of the null point) and lRl_R the length DB. Since the wire is uniform, RAD=ρlxAR_{AD} = \dfrac{\rho l_x}{A} and RDB=ρlRAR_{DB} = \dfrac{\rho l_R}{A}, so the ρ\rho and AA cancel and only lengths matter.
lx+lR=100 cml_x + l_R = 100\ \text{cm}
  • The wire is one metre long, so a null point at lxl_x cm from the left end leaves (100−lx)(100-l_x) cm on the right.
G=(lg100−lg)RG = \left(\dfrac{l_g}{100-l_g}\right)R
  • Kelvin's method for the resistance of a galvanometer. The galvanometer GG occupies one gap and a known RR the other; lgl_g is the length of wire on the side of the galvanometer and lr=100−lgl_r = 100 - l_g the length on the side of the resistance box.
X=RQPX = \dfrac{RQ}{P}
  • Post office box (a practical Wheatstone bridge). PP and QQ are the fixed ratio arms (each settable to 10, 100 or 1000 ohm), RR is the adjustable known arm (1 ohm to 5000 ohm) and XX the unknown fourth arm. The specific resistance of a wire of length LL and radius rr used as XX is ρ=X πr2L\rho = \dfrac{X\,\pi r^2}{L}.

Potentiometer

FormulaSymbols, units and conditions
I=ER+rVAB=IR=ERR+rI = \dfrac{E}{R+r} \qquad V_{AB} = IR = \dfrac{ER}{R+r}
  • Current through, and potential difference across, a potentiometer wire AB of resistance RR driven by an auxiliary cell of emf EE and internal resistance rr.
K=VABL=ERL(R+r)K = \dfrac{V_{AB}}{L} = \dfrac{ER}{L(R+r)}
  • Potential gradient KK: the potential difference per unit length of the potentiometer wire, in V m−1\text{V m}^{-1} (or V cm−1\text{V cm}^{-1}). It stays constant as long as EE and the circuit are unchanged. With a series (rheostat or box) resistance XX added, K=(ER+r+X)RLK = \left(\dfrac{E}{R+r+X}\right)\dfrac{R}{L}.
K=IRL=IσK = \dfrac{IR}{L} = I\sigma
  • The same gradient written from the current in the wire; σ=RL\sigma = \dfrac{R}{L} is the resistance per unit length of the wire, in Ω m−1\Omega\,\text{m}^{-1}.
VAC=Kli.e. VAC∝lV_{AC} = Kl \qquad \text{i.e.}\ V_{AC} \propto l
  • Principle of the potentiometer: the potential difference between two points of the wire is directly proportional to the length of wire between them, provided the wire has uniform cross section, carries a steady current, and stays at constant temperature.
E1E2=l1l2\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}
  • Comparison of emfs by connecting the two cells individually (Method I). l1l_1 and l2l_2 are the balancing lengths from A for cells of emf E1E_1 and E2E_2, since E1=Kl1E_1 = Kl_1 and E2=Kl2E_2 = Kl_2.
E1+E2=Kl1E1−E2=Kl2∴ E1E2=l1+l2l1−l2E_1 + E_2 = Kl_1 \qquad E_1 - E_2 = Kl_2 \qquad \therefore\ \dfrac{E_1}{E_2} = \dfrac{l_1+l_2}{l_1-l_2}
  • Comparison of emfs by the sum and difference (combination) method, E1>E2E_1 > E_2. l1l_1 is the balancing length in the sum mode and l2l_2 in the difference mode; the last step follows by componendo and dividendo.
r=R(l1l2−1)=R(l1−l2l2)r = R\left(\dfrac{l_1}{l_2} - 1\right) = R\left(\dfrac{l_1-l_2}{l_2}\right)
  • Internal resistance of a cell by potentiometer. l1l_1 is the balancing length with the cell on open circuit (key K2K_2 open, length corresponding to emf E1E_1) and l2l_2 the balancing length with the resistance box RR closed across the cell (length corresponding to terminal p.d. VV).
E1V=l1l2=R+rR\dfrac{E_1}{V} = \dfrac{l_1}{l_2} = \dfrac{R+r}{R}
  • The intermediate result of that derivation; useful directly when the question gives EE and VV instead of two lengths.
V1=l1l1+l2VV2=l2l1+l2VV_1 = \dfrac{l_1}{l_1+l_2}V \qquad V_2 = \dfrac{l_2}{l_1+l_2}V
  • Potentiometer as a voltage divider: a total voltage VV across AB divides between the two parts of the wire in the ratio of their lengths l1l_1 and l2l_2.

Galvanometer

FormulaSymbols, units and conditions
GIg=SIsIs=I−IgGI_g = SI_s \qquad I_s = I - I_g
  • A shunt SS (a low resistance) in parallel with a moving coil galvanometer of resistance GG: they have the same potential difference across them, and the main current II splits into IgI_g through the coil and IsI_s through the shunt.
S=(IgI−Ig)GS = \left(\dfrac{I_g}{I-I_g}\right)G
  • Shunt required so that a galvanometer of full-scale deflection current IgI_g reads up to a maximum current II — that is, so it works as an ammeter of range 00 to II.
S=Gn−1n=IIgS = \dfrac{G}{n-1} \qquad n = \dfrac{I}{I_g}
  • Shunt required to increase the current range nn times.
IsI=GS+G\dfrac{I_s}{I} = \dfrac{G}{S+G}
  • Fraction of the total current that passes through the shunt. (The fraction through the galvanometer is IgI=SS+G\dfrac{I_g}{I} = \dfrac{S}{S+G}.)
RA=SGS+G=GnR_A = \dfrac{SG}{S+G} = \dfrac{G}{n}
  • Resistance of the resulting ammeter — small, as it should be. An ideal ammeter has zero resistance and is connected in series.
X=VIg−GX = \dfrac{V}{I_g} - G
  • High resistance XX to be connected in series with the galvanometer to convert it into a voltmeter reading up to VV volt.
X=G(nv−1)nv=VVg=VIgGX = G(n_v - 1) \qquad n_v = \dfrac{V}{V_g} = \dfrac{V}{I_gG}
  • The same result written through the factor nvn_v by which the voltage range is increased, Vg=IgGV_g = I_gG being the voltage across the galvanometer at full-scale deflection.
RV=G+X=GnvR_V = G + X = Gn_v
  • Resistance of the resulting voltmeter — large, as it should be. An ideal voltmeter has infinite resistance and is connected in parallel.
IgI=S(G+X)+S\dfrac{I_g}{I} = \dfrac{S}{(G+X)+S}
  • Used when the only available shunt SS is too large to give the desired range: a resistance XX is added in series with the galvanometer coil, and the coil branch then has resistance G+XG+X.
E=αT+12βT2E = \alpha T + \dfrac{1}{2}\beta T^2
  • Thermo emf of a thermocouple whose cold junction is at 0 ∘C0\,^\circ\text{C} and hot junction at T ∘CT\,^\circ\text{C}; α\alpha and β\beta are the thermoelectric constants. The graph of EE against TT is a parabola. (Seebeck effect, from the "Do you know" box.)
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.5 subtopics
  1. 1Introduction
  2. 2Kirchhoff's Laws of Electrical Network4 board questions
  3. 3Wheatstone Bridge11 board questions
  4. 4Potentiometer18 board questions
  5. 5Galvanometer11 board questions
The whole chapter in brief

9.1 Introduction

  • Std XI dealt with the origin of electrical conductivity and with combining resistances in series and in parallel.
  • A circuit with several complex connections cannot always be reduced to one loop by series and parallel rules alone — for instance a Wheatstone network, where no two resistors are purely in series or purely in parallel.
  • Gustav Robert Kirchhoff (1824-1887) framed two rules that analyse any such network. This chapter is those two laws and three instruments built on them: the Wheatstone bridge, the potentiometer and the galvanometer.

9.2 Kirchhoff's laws — the vocabulary first

  • Junction: any point in an electric circuit where two or more conductors are joined together.
  • Loop (mesh): any closed conducting path in an electric network.
  • Branch: any part of the network that lies between two junctions.
  • The laws hold for a steady current in a network of resistors, and apply to both DC and AC circuits (for AC one uses instantaneous values).

First law (current law, junction law)

  • Statement: the algebraic sum of the currents at a junction in an electrical network is zero, ∑i=1nIi=0\sum_{i=1}^{n}I_i = 0, where IiI_i is the current in the ithi^{th} conductor at a junction having nn conductors.
  • Sign convention: arriving currents positive, leaving currents negative. For the six-conductor junction P of Fig. 9.2, I1−I2+I3+I4−I5−I6=0I_1 - I_2 + I_3 + I_4 - I_5 - I_6 = 0, i.e. I1+I3+I4=I2+I5+I6I_1+I_3+I_4 = I_2+I_5+I_6.
  • It is consistent with conservation of electric charge: charge arriving per unit time must leave per unit time, since charge cannot pile up at a junction.

Second law (voltage law, loop theorem)

  • Statement: the algebraic sum of the potential differences (products of current and resistance) and the electromotive forces in a closed loop is zero, ∑IR+∑E=0\sum IR + \sum E = 0.
  • Sign convention for a resistor: travelling along the conventional current, the p.d. is negative; travelling against it, positive.
  • Sign convention for a source: tracing within the source from the negative to the positive terminal, the emf is positive; from positive to negative, negative.
  • It is consistent with conservation of energy: the energy per unit charge supplied by the sources reappears as potential differences across the elements of the loop.

Steps the board expects in a Kirchhoff numerical

  • Choose some direction for each current.
  • Reduce the number of variables using the first law (write the third current as the sum of two).
  • Determine the number of independent loops.
  • Apply the voltage law to all the independent loops.
  • Solve the resulting equations simultaneously.
  • If a current comes out negative, the actual conventional current flows opposite to the direction chosen — say so, do not simply drop the sign.

9.3 Wheatstone bridge

  • Resistance changes with temperature, strain, humidity, displacement, liquid level and so on, so measuring resistance is a way of measuring those quantities. Measurable resistances run from a few milliohm to hundreds of megohm; a Wheatstone bridge is generally used for the range tens of ohm to hundreds of ohm.
  • Developed by Charles Wheatstone (1802-1875) to measure unknown resistances; also used to calibrate voltmeters and ammeters.
  • Four resistances PP, QQ, RR, SS form the quadrilateral ABCD; a cell of emf EE with a key sits between A and C (A at higher potential), and a galvanometer of resistance GG between B and D.
  • Balance means Ig=0I_g = 0, obtained by adjusting PP, QQ, RR, SS. Balancing condition: PQ=SR\dfrac{P}{Q} = \dfrac{S}{R}, i.e. the products of opposite arms are equal, PR=QSPR = QS.
  • Applications: measuring very low resistances precisely; measuring galvanometer resistance, capacitance, inductance and impedance; with an operational amplifier, measuring temperature, strain and similar physical parameters.

9.3.1 Metre bridge

  • A one-metre wire of uniform cross section stretched along a metre scale on a wooden board. Its ends are fixed under two L-shaped metallic strips, and a single central strip leaves two gaps — the left gap (unknown XX) and the right gap (resistance box RR).
  • One galvanometer terminal goes to the central strip C, the other carries the jockey J, which makes temporary contact with the wire. A cell, key and rheostat are connected between A and B.
  • The wire itself supplies the other two arms of the Wheatstone bridge, so XR=lxlR\dfrac{X}{R} = \dfrac{l_x}{l_R} and X=lxlRRX = \dfrac{l_x}{l_R}R, with lx+lR=100l_x + l_R = 100 cm.
Sources of error in a metre bridgeHow it is minimised
The cross section of the wire may not be uniformRepeat the experiment by interchanging XX and RR and take the mean
Contact resistance develops where the wire is soldered to the metallic strips, and is not accounted forInterchanging XX and RR compensates for the end resistances
The measurements of lxl_x and lRl_R may not be accurateAdjust RR so the null point lies in the middle one third of the wire, between 34 cm and 66 cm, where the percentage errors in lxl_x and lRl_R are minimum and nearly the same
Sliding the jockey scrapes the wire and spoils its uniformityTap the jockey on the wire, never slide it — it is only there to detect whether current flows in the central branch

Kelvin's method for the resistance of a galvanometer

  • The galvanometer whose resistance GG is wanted goes in one gap, a known resistance box RR in the other.
  • A suitable RR is chosen, the key is closed and the deflection is noted without touching the jockey; the rheostat is adjusted to give a deflection of about two thirds of the range.
  • The jockey is tapped at different points until a point D is found at which the deflection does not change on contact. Since the deflection is the same with and without contact between B and D, those two points are equipotential — the bridge is balanced.
  • With lgl_g the length of wire opposite the galvanometer and lr=100−lgl_r = 100 - l_g the length opposite the box, GR=lglr\dfrac{G}{R} = \dfrac{l_g}{l_r}, so G=(lg100−lg)RG = \left(\dfrac{l_g}{100-l_g}\right)R.
  • Note the difference from an ordinary metre bridge: here the galvanometer is in a gap, not in the central branch, and balance is a no-change in deflection rather than a zero deflection.

Post office box and strain gauge

  • A post office box is a practical Wheatstone bridge with three adjustable arms: ratio arms PP and QQ (each 10, 100 or 1000 ohm) and RR (1 ohm to 5000 ohm), with the unknown XX — usually a wire — as the fourth arm, and two tap keys K1K_1 and K2K_2. At balance X=RQPX = \dfrac{RQ}{P}, and ρ=Xπr2L\rho = \dfrac{X\pi r^2}{L} for a wire of length LL and radius rr.
  • A strain gauge is a resistor whose resistance (typically 30 to 3000 ohm) is proportional to the strain in it. It replaces the unknown arm; R1=R2R_1 = R_2 and R3R_3 is a variable resistor. With no force applied, the rheostat is adjusted until the voltmeter reads zero — this balance is the zero of the gauge. Stretching or compressing the gauge changes its resistance, unbalances the bridge, and the voltmeter reading measures the strain; more strain gives a larger reading.

9.4 Potentiometer

  • An ideal voltmeter would have infinite resistance and draw no current, so that it does not disturb the p.d. it is measuring. No real voltmeter can. A potentiometer draws no current from the source at balance and therefore acts as an ideal voltmeter; it is used for accurate measurement of potential difference and of emf.

9.4.1 Principle

  • Principle: the potential difference between two points on the wire is directly proportional to the length of wire between them, provided (i) the wire is of uniform cross section, (ii) the current through the wire is the same, and (iii) the temperature of the wire remains constant.
  • I=ER+rI = \dfrac{E}{R+r}, VAB=ERR+rV_{AB} = \dfrac{ER}{R+r}, and the potential gradient is K=VABL=ERL(R+r)K = \dfrac{V_{AB}}{L} = \dfrac{ER}{L(R+r)}.
  • Potential gradient: potential difference per unit length of wire. For a point C at distance ll from A, VAC=KlV_{AC} = Kl.
  • KK depends on the emf of the auxiliary (driving) cell, the resistance and length of the potentiometer wire, its internal and series resistances — that is, on the current through the wire and on the resistance per unit length. It does not depend on the cell being tested.
  • Measuring KK: balance a cell of known emf EE at length ll; then K=ElK = \dfrac{E}{l}. Equivalently, KK is the p.d. across the whole wire divided by its length.

9.4.2 Uses

  • Comparing emfs, cells connected individually: E1E2=l1l2\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}.
  • Comparing emfs, sum and difference method: E1+E2=Kl1E_1+E_2 = Kl_1, E1−E2=Kl2E_1-E_2 = Kl_2, hence E1E2=l1+l2l1−l2\dfrac{E_1}{E_2} = \dfrac{l_1+l_2}{l_1-l_2}.
  • Sum method: the negative terminal of the first cell is joined to the positive terminal of the second, so the emfs add to E1+E2E_1+E_2. Difference method: like terminals are joined together, so the emfs oppose and the effective emf is E1−E2E_1-E_2 (with E1>E2E_1>E_2). The difference connection is not a parallel combination of cells.
  • Internal resistance of a cell: r=R(l1−l2l2)r = R\left(\dfrac{l_1-l_2}{l_2}\right), where l1l_1 balances the emf (box open) and l2l_2 the terminal p.d. (box RR closed).
  • Practical applications: as a voltage divider, to change the output voltage of a supply continuously, the voltage dividing in the ratio of the lengths l1l_1 and l2l_2; as audio controls, the sliding faders and rotary knobs used for frequency attenuation and loudness control; as a motion sensor, when the slider is attached to a moving part, so that the change in potential (amplified) is calibrated in terms of displacement.
  • Internal resistance of a cell itself depends on the nature and concentration of the electrolyte, the separation between the electrodes, the area of the electrodes dipped in the electrolyte, and the temperature.

9.4.3 Potentiometer versus voltmeter

PotentiometerVoltmeter
Its resistance is effectively infiniteIts resistance is high but finite
It draws no current from the source at balanceIt draws some current from the source
The p.d. it measures equals the actual p.d.The p.d. it measures is less than the actual p.d.
High sensitivity; a p.d. of the order of 10−610^{-6} V can be measuredLow sensitivity; its least count is far coarser
It measures emf as well as p.d., and the internal resistance of a cellIt measures only terminal p.d., never emf or internal resistance
More accurateLess accurate
It does not give a direct readingIt gives a direct reading
It is not portableIt is portable
  • Merits, in the board's own words: more sensitive than a voltmeter; can measure both p.d. and emf, whereas a voltmeter draws current and so always reads terminal p.d.; very accurate, with a least count far better than a voltmeter's.
  • Demerits: it is not portable, and direct measurement of p.d. or emf is not possible — every reading is a balancing length that must then be converted.

9.5 Galvanometer

  • A galvanometer detects weak electric currents. Its coil is pivoted (or suspended) between the concave pole faces of a strong laminated horseshoe magnet; current through the coil deflects it, and the deflection is proportional to the current and is read off by a pointer on a scale.
  • The coil has moderate resistance (about 100 ohm) and a small current capacity (about 1 mA), so on its own it can neither measure large currents nor withstand them.

9.5.1 As an ammeter

  • Three changes needed: increase its effective current capacity; decrease its effective resistance (an ideal ammeter has zero resistance, since a finite GG in series would itself reduce the current being measured); protect it from damage by excessive current.
  • All three are achieved by a shunt SS, a low resistance in parallel with the galvanometer.
  • Uses of a shunt: it diverts a large part of the total current through an alternative path and so protects the instrument; it increases the range of the ammeter; it decreases the resistance between the points to which it is connected.
  • S=(IgI−Ig)GS = \left(\dfrac{I_g}{I-I_g}\right)G, and for an nn-fold increase in range, S=Gn−1S = \dfrac{G}{n-1} with n=IIgn = \dfrac{I}{I_g}. Fraction through the shunt: IsI=GS+G\dfrac{I_s}{I} = \dfrac{G}{S+G}.

9.5.2 As a voltmeter

  • Three changes needed: increase its voltage measuring capacity; increase its effective resistance (an ideal voltmeter has infinite resistance, since a low-resistance meter draws current and lowers the very p.d. being measured); protect it from excess applied p.d.
  • All three are achieved by a high resistance XX in series with the galvanometer.
  • V=IgX+IgGV = I_gX + I_gG, hence X=VIg−GX = \dfrac{V}{I_g} - G, and in terms of the range multiplier, X=G(nv−1)X = G(n_v-1) with nv=VIgGn_v = \dfrac{V}{I_gG}.
AmmeterVoltmeter
It measures currentIt measures potential difference
It is connected in seriesIt is connected in parallel
It is an MCG with low resistance (ideally zero)It is an MCG with high resistance (ideally infinite)
Smaller the shunt, greater the current measuredLarger its resistance, greater the p.d. measured
RA=SGS+G=GnR_A = \dfrac{SG}{S+G} = \dfrac{G}{n}RV=G+X=GnvR_V = G+X = Gn_v

Thermoelectricity (Seebeck effect)

  • A resistor converts electrical energy into thermal energy. The reverse conversion, thermal energy directly into electrical energy, was discovered by Seebeck and is called the thermoelectric effect.
  • If two different metals are joined to form a closed loop and the two junctions are kept at different temperatures, a small emf (thermo emf) is produced and current flows. The pair of dissimilar metals is a thermocouple.
  • Direction: in an antimony-bismuth couple the current flows from antimony to bismuth at the cold junction (the ABC rule); in a copper-iron couple it flows from copper to iron at the hot junction.
  • The effect is reversible — interchanging the hot and cold junctions reverses the current.
  • With the cold junction at 0 ∘C0\,^\circ\text{C} and the hot junction at T ∘CT\,^\circ\text{C}, E=αT+12βT2E = \alpha T + \dfrac{1}{2}\beta T^2, where α\alpha and β\beta are the thermoelectric constants; the graph of EE against TT is therefore a parabola.

Standard results worth memorising

  • At the balance point of a potentiometer, current is drawn from the auxiliary (driving) battery only — none from the cell being tested.
  • Kirchhoff's first law is a statement of conservation of charge; the second law, of conservation of energy.
  • If the current through the potentiometer wire is increased, KK increases, so the balancing length l=EKl = \dfrac{E}{K} decreases and the null point shifts towards A. If the current is decreased, KK decreases and the null point shifts away from A, towards B — and if KK falls below EL\dfrac{E}{L}, no null point is obtained at all.
  • A resistance ss shunted across one arm to balance a bridge: for arms 10, 10, 10 and 15 ohm, the 15 ohm arm must read 10 ohm, so 15s15+s=10\dfrac{15s}{15+s} = 10 gives s=30 Ωs = 30\ \Omega.
  • A ring or bangle connected in a gap at two diametrically opposite points contributes two halves in parallel, i.e. R4\dfrac{R}{4} where RR is the resistance of the whole ring.
  • A circular loop of total resistance RR tapped at two points a quarter of the circumference apart offers R4\dfrac{R}{4} and 3R4\dfrac{3R}{4} in parallel, i.e. 3R16\dfrac{3R}{16}.
  • A voltmeter of resistance RVR_V across a cell of emf EE and internal resistance rr reads V=ERVRV+rV = \dfrac{ER_V}{R_V+r}, always less than EE.

Traps that cost marks

  • Writing the balancing condition as PQ=RS\dfrac{P}{Q} = \dfrac{R}{S}. The textbook's labelling gives PQ=SR\dfrac{P}{Q} = \dfrac{S}{R}. Safest is to remember that the products of opposite arms are equal.
  • Forgetting that in the metre bridge lxl_x and lRl_R are lengths of the same wire, so ρ\rho and AA cancel; students sometimes try to substitute a resistivity that was never given.
  • Mixing up lgl_g in Kelvin's method: it is the length of wire on the side of the galvanometer, and 100−lg100-l_g is on the side of the box. Interchanging them inverts the answer.
  • In the internal resistance formula r=R(l1−l2l2)r = R\left(\dfrac{l_1-l_2}{l_2}\right), l1l_1 is the longer, open-circuit length. Getting l1l_1 and l2l_2 the wrong way round gives a negative rr.
  • Writing E1E2=l1−l2l1+l2\dfrac{E_1}{E_2} = \dfrac{l_1-l_2}{l_1+l_2} in the sum and difference method. The sum length l1l_1 is the larger one and belongs on top: E1E2=l1+l2l1−l2\dfrac{E_1}{E_2} = \dfrac{l_1+l_2}{l_1-l_2}.
  • Using X=VIgX = \dfrac{V}{I_g} and forgetting to subtract GG when converting a galvanometer into a voltmeter.
  • Substituting IgI_g in mA or μ\muA into S=(IgI−Ig)GS = \left(\dfrac{I_g}{I-I_g}\right)G while II is in A. Convert both to the same unit first.
  • Confusing n=IIgn = \dfrac{I}{I_g} (ammeter) with nv=VIgGn_v = \dfrac{V}{I_gG} (voltmeter); S=Gn−1S = \dfrac{G}{n-1} is a division, X=G(nv−1)X = G(n_v-1) a multiplication.
  • Dropping the minus sign in a Kirchhoff answer. A negative current is a correct answer with a physical meaning — the current flows opposite to the assumed direction — and the board expects that sentence.
  • Saying the potentiometer "draws a very small current" at balance. At balance it draws exactly zero current from the cell being tested; that is the whole point.
  • Sliding the jockey instead of tapping it, and being asked why not: the wire gets scraped, its cross section is no longer uniform, and the null point becomes unreliable.
  • Claiming a voltmeter can measure emf. It measures terminal p.d. only, because it always draws some current.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 2 have come up in two or more papers.2
  1. 1
    Balancing condition of a Wheatstone's bridge Repeats
    Obtain the balancing condition in the case of Wheatstone's bridge.

    Asked in: October 2008; July 2024 (Q.19, 3 marks)

    Show the derivation
    Construction:
    Step 1
    Four resistances P, Q, R and S are connected to form a quadrilateral ABCD as shown in the figure.
    Step 2
    A battery of emf E along with a key is connected between the points A and C such that point A is at higher potential with respect to the point C.
    Step 3
    A galvanometer of internal resistance G is connected between points B and D.
    Working:
    Step 1
    When the key is closed, current I flow through the circuit. It divides into I1I_1 and I2I_2 at point A. I1I_1 is the current through P and I2I_2 is the current through S.
    ∴I=I1+I2\therefore I=I_1+I_2
    Step 2
    The current I1I_1 gets divided at point B. Let IgI_g be the current flowing through the galvanometer. The currents flowing through Q and R are (I1−Ig)(I_1-I_g) and (I2+Ig)(I_2+I_g) respectively.
    Step 3
    Consider the loop ABDA, applying Kirchhoff's voltage law in the clockwise sense shown in the loop.
    ∴−I1P−IgG+I2S=0....(1)\therefore -I_1P-I_gG+I_2S=0\qquad....(1)
    Step 4
    Now consider loop BCDB, applying Kirchhoff's voltage law in the clockwise sense as shown in the loop,
    −(I1−Ig)Q+(I2+Ig)R+IgG=0....(2)-(I_1-I_g)Q+(I_2+I_g)R+I_gG=0\qquad....(2)
    Step 5
    The bridge is said to be balanced, when the current passing through the galvanometer is zero, i.e., Ig=0I_g=0. This condition can be obtained by adjusting the values of P, Q, R and S.
    Step 6
    Substituting Ig=0I_g=0 in equation (1),
    −I1P+I2S=0-I_1P+I_2S=0
    ∴I1P=I2S....(3)\therefore I_1P=I_2S\qquad....(3)
    Substituting Ig=0I_g=0 in equation (2),
    −I1Q+I2R=0-I_1Q+I_2R=0
    ∴I1Q=I2R....(4)\therefore I_1Q=I_2R\qquad....(4)
    Dividing equation (3) by equation (4),
    PQ=SR\dfrac{P}{Q}=\dfrac{S}{R}
    This is the balancing condition for Wheatstone bridge.
  2. 2
    Conversion of a galvanometer into a voltmeter Repeats
    Explain how a moving coil galvanometer can be converted into a voltmeter. Derive the necessary formula.

    Asked in: March 2008; March 2017

    Show the derivation
    Step 1
    To use a M.C.G as a voltmeter, a high resistance is connected in series with the M.C.G.
    Step 2
    A very high resistance X is connected in series with the galvanometer for this purpose as shown in figure.
    Step 3
    If V is the voltage to be measured, then
    V=IgX+IgGV=I_gX+I_gG
    ∴IgX=V−IgG\therefore I_gX=V-I_gG
    ∴X=VIg−G....(1)\therefore X=\dfrac{V}{I_g}-G\qquad....(1)
    where IgI_g is the current flowing through the galvanometer.
    Step 4
    If voltage V is nvn_v times voltage VgV_g (voltage across galvanometer) then,
    V=nvVg=nv(IgG)V=n_vV_g=n_v(I_gG)
    Using this in equation (1),
    X=G(nv−1).X=G(n_v-1).

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