Current Electricity — JEE Mains Physics Formulas and Traps

Current, Drift Velocity and Current Density

Charge, current and the area under an I–t graph

Charge and current

I = \frac{dq}{dt}, \qquad q = \int_{t_1}^{t_2} I\,dt, \qquad n = \frac{I}{e}

Drift velocity, mobility and current density

Drift and current density

I = neAv_d, \qquad v_d = \frac{eE\tau}{m} = \mu E, \qquad J = nev_d = \sigma E

Common traps

Use the limits the question gives

"From t = 1 s to t = 2 s" means the lower limit is 1, not 0. Integrating from zero adds the charge of the first second and lands on a wrong option.

mAh is not coulombs

Multiply mAh by 3.6 to get coulombs. Then energy = charge × voltage. Forgetting the 3600 s in an hour gives an answer smaller by a factor of 3600.

Least current comes from dI/dt, not dq/dt

When the charge is given as q(t), differentiate once to get I and once more to find where I is least. Setting dq/dt = 0 finds where the current is zero instead.

mm² is 10⁻⁶ m²

Areas are usually given in mm². Leaving them in mm² makes the drift speed a million times too small.

Electrons drift towards higher potential

Conventional current runs from high to low potential. Electrons are negative, so they drift the other way, against the field. A statement that electrons drift from higher to lower potential describes conventional current, not the electrons.

At fixed voltage the area does not matter

Doubling the area at the same voltage doubles the current but leaves the drift speed, (eτ/m)(V/l), unchanged. Only at a fixed current does a larger area mean a smaller drift speed.

Resistance, Resistivity and Temperature

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Resistivity and the shape of a conductor

Resistance from shape

R = \frac{\rho l}{A}, \qquad A_{\text{hollow}} = \pi\left(r_2^{2} - r_1^{2}\right), \qquad \sigma = \frac{1}{\rho}

Stretching, melting and redrawing a wire

Constant-volume stretching

R' = n^{2}R \ (\text{length} \times n), \qquad R \propto \frac{1}{r^{4}}, \qquad \frac{\Delta R}{R} \approx 2\frac{\Delta l}{l}

Resistance and temperature

Temperature dependence

R_T = R_0\left(1 + \alpha\,\Delta T\right), \qquad t = \frac{R_t - R_0}{R_{100} - R_0} \times 100\ ^{\circ}\text{C}

Common traps

Radius, not diameter

Questions give diameters. The area uses radii, so halve them first; using the diameter makes the area four times too large.

Resistivity does not change with shape

Doubling the length doubles R but leaves ρ alone. ρ changes only with the material and the temperature.

Which faces?

For a block, the current's length is the distance between the two faces it enters and leaves by, and A is the area of those faces. Swapping the two gives an answer off by a large factor.

R goes as n², not n

Stretching to n times the length also thins the wire n times in area. Scaling R by n alone is the most common wrong option.

2Δl/l is only for small changes

For a 0.5% stretch, 2Δl/l is fine. For 20% or more, square the factor exactly; the estimate misses by several per cent.

"By twice" is three times

A length increased by twice its value is 3l, not 2l. Read the wording before squaring.

Know the reference temperature

α is defined with R₀ at 0 °C. If the readings are at 10 °C and 30 °C, taking the 10 °C value as R₀ gives a slightly different α, and the options are often built on one choice or the other.

Kelvin at the end

A temperature change is the same in °C and K, so work in °C and add 273 only when the answer is asked in kelvin.

A hotter wire draws less current

At a fixed voltage, a rise in R means a fall in current. Treating the current as fixed and the voltage as changing inverts the ratio.

Equivalent Resistance: Series, Parallel and Symmetry

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Series, parallel and bent wires

Series, parallel, loops

R_s = \sum R_i, \qquad \frac{1}{R_p} = \sum \frac{1}{R_i}, \qquad R_{\text{loop}} = R\,x(1 - x)

Redrawing networks: shorts and symmetry

The two moves

V_P = V_Q \ \Rightarrow\ I_{PQ} = 0, \qquad R_{\text{eq}} = \frac{V_{\text{in}} - V_{\text{out}}}{I_{\text{total}}}

Common traps

A cut wire divides twice

Cutting into n pieces makes each R/n, and putting them in parallel divides by n again. The answer is R/n², not R/n.

Arcs go by length

On a bent wire the resistance of an arc is in proportion to its length. Equal angles at the centre of a circle mean equal resistances; unequal sides of a shape do not.

Check every option numerically

Combination questions offer several plausible drawings. Compute each one; the right one is often not the most symmetric.

A missed short

A plain wire drawn along the edge of a figure is easy to overlook. It merges two points and can remove a whole resistor. Label the nodes before writing any formula.

A crossing is not a junction

Wires that cross without a dot, or with a hump, are separate. Joining them changes the network and the answer.

Symmetry must be about the input–output line

A network can look symmetric in many ways, but only symmetry with respect to where the current enters and leaves gives equal potentials.

Kirchhoff's Laws and the Wheatstone Bridge

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Current and voltage dividers

Dividers

V_1 = V\frac{R_1}{R_1 + R_2}, \qquad I_1 = I\frac{R_2}{R_1 + R_2}

The balanced Wheatstone bridge

Balance condition

\frac{P}{Q} = \frac{R}{S} \ \Rightarrow\ I_{\text{middle}} = 0, \qquad R_{\text{eq}} = \frac{(P + Q)(R + S)}{P + Q + R + S}

Kirchhoff's laws and node potentials

Junction equation at a node x

\sum_i \frac{V_x - V_i}{R_i} = 0, \qquad \sum \varepsilon = \sum IR

Common traps

Current divides inversely

In parallel, the larger share of current goes through the SMALLER resistor. Putting a resistor's own value on top of the fraction gives the other branch's current.

A rating is not the working voltage

A 200 V bulb in a 100 V circuit does not have 200 V across it. Use the rating only to find R, then let the circuit decide the voltage.

A load lowers the output

Anything connected across part of a divider lowers that part's resistance and so its voltage. Using the unloaded value is the trap option.

Pair the arms correctly

The ratio compares the two arms on the same side of the middle arm in each branch: P/Q = R/S, where P and R both touch the same input point. Cross-pairing gives a wrong unknown.

The middle arm's value is irrelevant at balance

At balance the middle arm carries nothing, so its resistance never enters the answer. A question that gives it is testing whether you notice.

Check before assuming balance

Five resistors in a diamond look like a bridge, but only equal ratios make it balanced. If the ratios differ, use Kirchhoff's laws.

The sign of a cell depends on the direction you cross it

Crossing a cell from − to + is a rise of ε, from + to − a drop. The direction of the current through the cell does not matter for this sign; it matters only for the IR term.

A negative current is not a mistake

If a current comes out negative, it flows opposite to the arrow you drew. Keep the size; flip the direction.

One equation per unknown node

With node potentials you need as many junction equations as unknown potentials, no more. Writing loop equations on top of them only adds algebra.

Cells: EMF, Internal Resistance and Combinations

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EMF, internal resistance and terminal voltage

A real cell

I = \frac{\varepsilon}{R + r}, \qquad V = \varepsilon \mp Ir, \qquad P_{\max} = \frac{\varepsilon^{2}}{4r}\ (R = r)

Cells in series, parallel and opposition

Two cells in parallel

\varepsilon_{\text{eq}} = \frac{\varepsilon_1/r_1 + \varepsilon_2/r_2}{1/r_1 + 1/r_2}, \qquad r_{\text{eq}} = \frac{r_1r_2}{r_1 + r_2}

Common traps

Charging adds Ir

When a cell is pushed backwards by a stronger one, its terminal voltage is ε + Ir, higher than its emf. Using ε − Ir for the weaker cell in opposition gives the wrong sign.

Maximum power is at R = r, not R = 0

A short circuit gives the largest current but no power in the load. The power in R peaks when R matches r.

Include r in the total

The current is ε/(R + r). Dividing ε by R alone forgets the cell's own resistance.

Weight the emfs by 1/r

The emf of cells in parallel is not the plain average. A 6 V cell with 2 Ω and a 4 V cell with 1 Ω give 14/3 V, not 5 V.

A reversed cell still adds its resistance

Reversing a cell flips the sign of its emf term only. Its internal resistance stays in the combined internal resistance exactly as before.

Equivalent emf lies between the emfs

For two cells facing the same way in parallel, the combined emf is between the two emfs, and the combined internal resistance is smaller than either. An answer outside that range signals a slip.

Meters, Meter Bridge and Potentiometer

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Real voltmeters and ammeters

Loaded reading

V_{\text{read}} = V\,\frac{R \parallel R_V}{R \parallel R_V + R_{\text{rest}}}, \qquad I_A = I\,\frac{S}{S + R_A}

The meter bridge

Balance

\frac{P}{Q} = \frac{l}{100 - l}

The potentiometer

Balance and internal resistance

\varepsilon = kl, \qquad k = \frac{V_{\text{wire}}}{L}, \qquad r = R\,\frac{l_1 - l_2}{l_2}

Common traps

A real voltmeter reads low

It lowers the resistance it sits across, and so the voltage there. Use the ideal divider value only when the question says the meter is ideal.

Higher meter resistance is better

A voltmeter with a much larger resistance than the resistor it measures barely changes the circuit. Of two meters, the one with the higher resistance gives the truer reading.

The slope is not R

If the voltmeter sits across R and the ammeter measures the total current, V/I gives R in parallel with the meter's own resistance. Only an ideal voltmeter makes it R.

Measure from P's end

l is the length on the same side as the gap on top of the fraction. Measuring from the other end swaps the ratio.

Which way does a shunt move the null point?

A shunt lowers that gap's resistance, so its length share shrinks and the null point moves towards that gap's end. Picture the ratio before writing numbers.

Wire properties drop out

Changing the wire's radius or material, or its resistance per cm, changes nothing at balance. Only a non-uniform wire or an end correction shifts the null point.

The wire gets only its share

The gradient uses the voltage across the wire, not the driver's full emf. A series resistance or the driver's internal resistance takes part of it.

Shunted means terminal voltage

A shunted cell delivers current, so its balance gives ε − Ir, not ε. Only the open-circuit balance gives the emf.

Longer wire, more sensitive

Sensitivity improves as the gradient falls, so a longer wire or a smaller current helps. A larger current makes it worse.

Electrical Power and Heating

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Power ratings, bulbs and heaters

From a rating

R = \frac{V_0^{2}}{P_0}, \qquad P = P_0\left(\frac{V}{V_0}\right)^{2}, \qquad R_s = \frac{V - V_0}{P_0/V_0}

Joule heating, losses and efficiency

Joule's law

H = I^{2}Rt = VIt = \frac{V^{2}t}{R}, \qquad Q = mc\,\Delta T + mL

Common traps

The resistance stays, the power changes

Off its rated supply a device keeps its resistance, not its wattage. Find R from the rating first, then the new power.

In series the lower rating glows more

A 40 W bulb has more resistance than a 60 W bulb of the same voltage, so with the same current it dissipates more. In parallel the order flips.

Power goes as the square

A 10% fall in current or voltage cuts the power by 19%, not 10%. Square the factor.

In parallel the smaller resistor heats more

Resistors in parallel share a voltage, so heat goes as V²/R: the smaller resistor gets the larger share. Using I²R with a single current gets it backwards.

"Whole circuit" includes r

The power of the whole circuit is εI, which counts the heat inside the cell. The power in the external resistors alone is smaller.

Warm before you melt

Ice below 0 °C must first be warmed to 0 °C (mcΔT) and then melted (mL). Leaving out either term gives a time that is too short.

Capacitors and Inductors in DC Circuits

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Capacitors in steady state

Steady state

I_{\text{C branch}} = 0, \qquad Q = CV_C, \qquad U = \tfrac{1}{2}CV_C^{2}

Just after switching and long after

The two moments

t = 0^{+}: \ L \to \text{open},\ C \to \text{wire}; \qquad t \to \infty: \ L \to \text{wire},\ C \to \text{open}

Charging, discharging and the time constant

Exponential change

q = q_0e^{-t/RC}, \qquad i = \frac{E}{R}\left(1 - e^{-tR/L}\right), \qquad t_{1/n} = \tau\ln n

Common traps

The series resistor drops nothing

With no current in the capacitor branch, a resistor in that branch has no voltage across it. Subtracting an IR drop there gives a wrong capacitor voltage.

Leave the capacitor branch out of the equivalent resistance

In steady state a capacitor branch is an open circuit. Including its resistor in the equivalent resistance changes the current everywhere else.

The capacitor voltage is a difference of potentials

Find the potential at each plate's node from the rest of the circuit and subtract. Do not assume the capacitor has the battery's full emf.

Which element is the wire, and when

At the first instant the capacitor is the wire and the inductor is the gap; long after it is the reverse. Swapping them gives the other moment's answer, which is usually an option.

A real inductor keeps its resistance

Long after switching, an inductor with its own resistance acts as that resistor, not as a plain wire.

Nothing jumps

An inductor's current and a capacitor's voltage just after a switch equal their values just before it. Use that, not the steady-state rule, for the first instant.

Energy decays twice as fast

Energy goes as q², so it falls as e^(−2t/RC). The energy halves in half the time the charge takes to halve.

Growth uses 1 − e^(−t/τ)

For charging or growing current, set 1 − e^(−t/τ) equal to the fraction asked, not e^(−t/τ). The two give different times except at one half.

ln, not log₁₀

The time is τ ln n with the natural logarithm. A value printed as log 3 = 1.1 is really ln 3: log₁₀ 3 is only about 0.48.

Current Electricity formulas

Current, Drift Velocity and Current Density

Charge, current and the area under an I–t graph

Drift velocity, mobility and current density

Common traps

Resistance, Resistivity and Temperature

Resistivity and the shape of a conductor

Stretching, melting and redrawing a wire

Resistance and temperature

Common traps

Equivalent Resistance: Series, Parallel and Symmetry

Series, parallel and bent wires

Redrawing networks: shorts and symmetry

Common traps

Kirchhoff's Laws and the Wheatstone Bridge

Current and voltage dividers

The balanced Wheatstone bridge

Kirchhoff's laws and node potentials

Common traps

Cells: EMF, Internal Resistance and Combinations

EMF, internal resistance and terminal voltage

Cells in series, parallel and opposition

Common traps

Meters, Meter Bridge and Potentiometer

Real voltmeters and ammeters

The meter bridge

The potentiometer

Common traps

Electrical Power and Heating

Power ratings, bulbs and heaters

Joule heating, losses and efficiency

Common traps

Capacitors and Inductors in DC Circuits

Capacitors in steady state

Just after switching and long after

Charging, discharging and the time constant

Common traps

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