Maharashtra HSC · Class XII · Physics

HSC Physics Revision Hub

Triage notes for the board exam. Use the chapter switcher below, or jump to the star-marked Formula Cheat Sheet for the whole syllabus in one place. Chapters 1, 2, 3, 4, 8, 9, 10 and 11 are written out in full; open the sheet and chapters to revise.

Kirchhoff's laws

Two conservation laws that govern every current circuit. Together they determine the current in every branch of any network.

Junction (node) rule - conservation of charge

At any junction, the algebraic sum of currents meeting there is zero: currents flowing in are balanced by currents flowing out.

\sum I_{\text{in}} = \sum I_{\text{out}} \qquad\text{or}\qquad \sum_{\text{junction}} I = 0

Loop (mesh) rule - conservation of energy

Around any closed loop, the algebraic sum of all the EMFs equals the algebraic sum of all the IR drops. The net change in potential going around a loop is zero.

\sum \varepsilon = \sum I R \qquad\text{or}\qquad \sum \Delta V = 0

Sign convention

Fix one direction of traverse around the loop. Take a current as positive when it flows in your direction of travel (a fall, $V$ drops by $IR$ ); take an EMF as positive when going from its negative to its positive terminal (a rise, gained $\varepsilon$ ). Apply the rule to enough independent loops and junctions to solve for every unknown current.

Wheatstone bridge

Four resistances $P, Q, S, R$ form a diamond $ABCD$ with a galvanometer across the diagonal $BD$ and a battery and key across the diagonal $AC$ . When the bridge is balanced the galvanometer reads zero.

Balance condition - null deflection, Ig = 0

The bridge is balanced when no current flows through the galvanometer ( $I_g = 0$ ). Then the junctions $B$ and $D$ are at the same potential, and the ratio of the arms obeys

\frac{P}{Q} = \frac{R}{S}

Derivation (from KCL and KVL)

With $I_g = 0$ , the junction rule at $B$ and $D$ shows the same current flows through $P$ and $Q$ (call it $I_1$ ), and the same current flows through $R$ and $S$ (call it $I_2$ ). No current enters or leaves the galvanometer, so nothing branches off.

Apply the loop rule to loop $A - B - D - A$ , going through $P$ and $R$ . The only drops are across the two arms (the galvanometer carries no current), so

I_1 P - I_2 R = 0 \qquad\Rightarrow\qquad I_1 P = I_2 R \qquad\ldots(1)

Apply the loop rule to loop $B - C - D - B$ , going through $Q$ and $S$ .

I_1 Q - I_2 S = 0 \qquad\Rightarrow\qquad I_1 Q = I_2 S \qquad\ldots(2)

Divide equation (1) by equation (2). The common currents $I_1, I_2$ cancel to give the balance condition

\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \qquad\Rightarrow\qquad \frac{P}{Q} = \frac{R}{S}

Applications

If three resistances are known, the fourth can be found. The bridge gives very accurate measurement because the reading is taken at null deflection, so the galvanometer draws no current from the unknown. It is the idea behind the meter bridge below.

Meter bridge

A meter bridge is a practical Wheatstone bridge. Two gaps - an unknown resistance $X$ and a known resistance box $R$ - sit between L-shaped metal strips, joined by a one-metre uniform wire $AC$ . A jockey slides along the wire; at the null point the galvanometer reads zero.

Principle and derivation

The two arms $P$ and $Q$ of a Wheatstone bridge are replaced by two segments of the same uniform wire. Resistance of a wire is $R = \rho l / A$ , so for one uniform wire it is proportional to length. Hence the arm ratio equals the length ratio

\frac{P}{Q} = \frac{l_x}{l_r}

Applying the Wheatstone balance ( $P/Q = R/S$ ) with $S = R$ (the resistance box) and $P = X$ (the unknown),

\frac{X}{R} = \frac{l_x}{l_r} \qquad\Rightarrow\qquad X = R\,\frac{l_x}{l_r}

where $l_x$ is the balancing length on the unknown side and $l_r$ the balancing length on the resistance-box side, with $l_x + l_r = 100 \,\text{cm}$ .

End errors and how to reduce them

Sources - non-uniformity or uneven thickness of the wire, high or loose contact resistance at the jockey and the end terminals, residual resistance of the metal strips and connecting leads (so the balance point is never exactly at the true end), and a slightly inaccurate scale. The net effect is an "end correction" that shifts the apparent null length.

Minimizing - use a single straight, uniform wire of constant thickness with good, sharp, low-resistance contacts and thick copper strips; slide the jockey gently and take readings near the middle of the wire where the balance is sharpest; repeat each reading several times and average. Best of all, interchange $X$ and $R$ in the two gaps and repeat; then average the two results, which largely cancels the end correction because the error reverses sign when the arms are swapped.

Potentiometer

A long uniform wire $AB$ carries a steady current from a driver battery $E$ through a key and a rheostat. A secondary circuit - the cell under test, a galvanometer and a sliding jockey - taps off a balancing length at which the galvanometer reads zero.

Potential gradient

The drop in potential per unit length of the wire. If $V$ is the potential difference across the whole wire and $L$ its length,

K = \frac{V}{L} \qquad [\text{V m}^{-1}]

At the balance point no current flows in the secondary, so the EMF of the cell just equals the fall across the balancing length: $E = K\,l$ .

Comparing two EMFs

For cells of EMF $E_1$ and $E_2$ with balancing lengths $l_1$ and $l_2$ , since $E \propto l$ for constant current,

\frac{E_1}{E_2} = \frac{l_1}{l_2}

Sum and difference method

Connect the two cells in series so they aid, giving resultant $E_1 + E_2$ and balancing length $l_1 + l_2$ ; then connect them in opposition, giving $E_1 - E_2$ and length $l_1 - l_2$ . Dividing,

\frac{E_1 + E_2}{E_1 - E_2} = \frac{l_1 + l_2}{l_1 - l_2}

Internal resistance

With a cell of EMF $E$ and internal resistance $r$ :

Measure the balance when the cell drives no external current: $E = K l_1$ (its true EMF).
Close the cell across a known resistance box $R$ . Current flows, the terminal voltage drops to $V = K l_2$ .

Using $V = IR$ and $E = I(R + r)$ ,

K l_2 = \frac{K l_1\, R}{\,R + r\,} \qquad\Rightarrow\qquad r = R\left(\frac{l_1}{l_2} - 1\right)

Advantages over a voltmeter

The potentiometer is a null method: at balance it draws no current from the cell, so it measures the true EMF, unaffected by the source's internal resistance and without disturbing the circuit. It gives very high accuracy and can measure very small or slowly varying potentials across a wide range. A voltmeter, by contrast, has finite resistance, always draws some current and reads terminal potential difference rather than true EMF, limiting its accuracy.

Galvanometer conversion

A galvanometer is a sensitive current detector, specified by its resistance $G$ and its full-scale deflection current $I_g$ . It is converted into a practical meter by adding one shunt or multiplier resistor.

Galvanometer into an ammeter (range I) - low shunt S in parallel

To read a current $I$ (greater than $I_g$ ), a small shunt resistance $S$ is connected in parallel. The galvanometer carries $I_g$ and the shunt the rest, $I - I_g$ . Because they are in parallel their potential differences are equal,

I_g G = (I - I_g)\, S \qquad\Rightarrow\qquad S = \frac{I_g\, G}{I - I_g}

Symbols: $I_g$ full-scale current of the galvanometer, $G$ its resistance, $I$ the new full-scale range of the ammeter. An ammeter must have low resistance and is connected in series.

Galvanometer into a voltmeter (range V) - high multiplier R in series

To read a voltage $V$ , a large series resistance $R$ is added so that at full-scale deflection the current through the combination is just $I_g$ ,

V = I_g (G + R) \qquad\Rightarrow\qquad R = \frac{V}{I_g} - G

Symbols: $V$ the new full-scale voltage, $I_g$ full-scale galvanometer current, $G$ galvanometer resistance. A voltmeter must have high resistance and is connected in parallel.

Quick reference table

Quantity / law	Symbol	SI unit	Relation
Charge	Q	coulomb (C)	Q = I t
Current	I	ampere (A)	I = Q / t
EMF	ε	volt (V)	ε = W / q
Potential difference	V	volt (V)	V = W / q
Ohm's law	R	ohm (Ω)	V = I R
Resistance	R	ohm (Ω)	R = ρl / A
Resistivity (material)	ρ	ohm metre (Ω m)	ρ = R A / l
Conductivity	σ	siemens per metre (S/m)	σ = 1 / ρ
Series resistors	R_eq	ohm (Ω)	R_eq = R₁ + R₂ + …
Parallel resistors	R_eq	ohm (Ω)	1/R_eq = 1/R₁ + 1/R₂ + …
Current density	j	A/m²	j = I / A
Drift velocity	vᵇ	m/s	vᵇ = I / (n A e)
Power	P	watt (W)	P = V I = I²R = V²/R
Potentiometer gradient	K	V/m	K = V / L
Wheatstone balance	-	-	P / Q = R / S (Iₗ = 0)
Meter bridge	X	ohm (Ω)	X = R · l_x / l_r
Internal resistance	r	ohm (Ω)	r = R (l₁/l₂ − 1)

e = 1.6 × 10⁻¹⁹ Ccharge of an electron

n = ~10²⁹ m⁻³free electrons per unit volume in a metal

ρ = Cu ≈ 1.7 × 10⁻⁸ Ω mresistivity of copper (reference)