Current Electricity
Ohm's law, resistance, combinations, Kirchhoff's laws & measuring instruments
Current, Ohm's Law and Resistance
Electric Current, Drift Velocity and Ohm's LawTopic 1
Electric current is the rate of flow of charge, $I = \dfrac{q}{t}$, measured in amperes. Although conventional current is taken in the direction of positive-charge flow, in metals it is actually electrons drifting the other way. In the absence of a field, free electrons move randomly with no net flow; switching on a field gives them a small steady drift velocity $v_d$ superimposed on this random motion.
The current links to drift velocity through $I = neAv_d$, where $n$ is the number of free electrons per unit volume, $A$ the cross-section and $e$ the electronic charge. A surprising NEET fact emerges here: the drift velocity is tiny (a fraction of a millimetre per second), yet bulbs light instantly because the electric field is established along the wire at nearly the speed of light, setting all electrons moving together.
Ohm's law states that, at constant temperature, the current through a conductor is proportional to the potential difference across it: $V = IR$, where $R$ is the resistance in ohms. Conductors obeying this are ohmic; devices like diodes and filament bulbs are non-ohmic, with a curved $V$–$I$ graph, a distinction NEET likes to test through graphs.
Resistance arises because drifting electrons collide with the lattice. It depends on the material and geometry: $R = \rho\dfrac{L}{A}$, where $\rho$ is the resistivity, a property of the material alone. Stretching a wire or using a longer, thinner one raises its resistance. For metals, resistivity increases with temperature (more lattice vibrations), while for semiconductors it decreases — an important contrast for NEET.
| Quantity | Relation |
|---|---|
| Current | $I = q/t = neAv_d$ |
| Ohm's law | $V = IR$ |
| Resistance | $R = \rho\dfrac{L}{A}$ |
A current of $2\ \text{A}$ flows for $5\ \text{minutes}$. How much charge passes a point in the wire?
Show solution
$q = It = 2\times(5\times60) = 2\times300 = 600\ \text{C}$.
A wire of resistance $R$ is stretched to twice its length (volume constant). Find its new resistance.
Show solution
Doubling the length halves the area, so $R = \rho L/A$ becomes $\rho(2L)/(A/2) = 4\rho L/A = 4R$. The resistance becomes four times.
Electric current is the rate of flow of:
The resistance of a wire is given by:
Drift velocity of electrons in a typical wire is of the order of:
With rising temperature, the resistivity of a metal:
A device whose $V$–$I$ graph is a straight line through the origin is:
NEET tip: $I = neAv_d$ links current and drift. When a wire is stretched at constant volume, $R \propto L^{2}$. Metals: resistivity rises with $T$; semiconductors: it falls.
Combination of Resistors and EMFTopic 2
Resistors are combined to control current. In series, the same current flows through each and the resistances add: $R_s = R_1 + R_2 + \dots$, giving a value larger than any single resistor. In parallel, the same voltage appears across each and the reciprocals add: $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$, giving a value smaller than the smallest resistor. (Note this is exactly opposite to how capacitors combine.)
A source like a cell does work to drive charge around a circuit; the energy supplied per unit charge is its electromotive force (EMF), $\varepsilon$. EMF is not a force but a potential — the voltage the cell would give with no current drawn. Every real cell also has a small internal resistance $r$, so when it delivers current $I$ the usable terminal voltage drops to $V = \varepsilon - Ir$.
This explains why a car's headlights dim when the starter motor runs: the large current causes a large $Ir$ drop, lowering the terminal voltage. For a cell driving an external resistance $R$, the current is $I = \dfrac{\varepsilon}{R + r}$ — a standard NEET starting equation. The terminal voltage equals the EMF only when no current flows (open circuit).
Cells too can be combined. In series the EMFs add (useful for higher voltage), while in parallel the EMF stays the same but the combination can supply more current with reduced internal resistance. Maximum power is delivered to the external load when $R = r$, the impedance-matching condition occasionally asked in NEET.
| Quantity | Relation |
|---|---|
| Series resistance | $R_s = R_1 + R_2 + \dots$ |
| Parallel resistance | $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ |
| Terminal voltage | $V = \varepsilon - Ir$ |
| Circuit current | $I = \dfrac{\varepsilon}{R + r}$ |
Two resistors of $4\ \Omega$ and $4\ \Omega$ are connected in parallel. Find the equivalent resistance.
Show solution
$\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$, so $R_p = 2\ \Omega$ — half the value, as expected for equal parallel resistors.
A cell of EMF $6\ \text{V}$ and internal resistance $0.5\ \Omega$ drives a $5.5\ \Omega$ resistor. Find the current and terminal voltage.
Show solution
$I = \dfrac{\varepsilon}{R + r} = \dfrac{6}{5.5 + 0.5} = 1\ \text{A}$. Terminal voltage $V = \varepsilon - Ir = 6 - 1\times0.5 = 5.5\ \text{V}$.
Resistors in series have an equivalent resistance equal to:
The terminal voltage of a cell delivering current is:
The EMF of a cell equals its terminal voltage when:
The current in a circuit with external $R$ and internal $r$ is:
Maximum power is delivered to the load when:
NEET tip: Resistors add in series and combine reciprocally in parallel (opposite to capacitors). Always include internal resistance: $I = \varepsilon/(R+r)$, $V = \varepsilon - Ir$.
Kirchhoff's Laws, Heating and Instruments
Kirchhoff's Laws and Electrical PowerTopic 3
Simple series–parallel reduction fails for complex networks with multiple cells, so we use Kirchhoff's two laws. The junction rule (KCL) states that the total current entering a junction equals the total current leaving it — a statement of charge conservation. The loop rule (KVL) states that the algebraic sum of potential changes around any closed loop is zero — a statement of energy conservation. Applying them with a consistent sign convention solves almost any circuit.
A classic application is the Wheatstone bridge, four resistors in a diamond with a galvanometer across the middle. The bridge is balanced (no galvanometer current) when $\dfrac{P}{Q} = \dfrac{R}{S}$. At balance the central branch carries no current, so it can be removed — a trick that simplifies many NEET network problems. The metre bridge is a practical version used to measure an unknown resistance.
When current flows through a resistor, electrical energy converts to heat — Joule heating. The power dissipated is $P = VI = I^{2}R = \dfrac{V^{2}}{R}$, and the heat produced in time $t$ is $H = I^{2}Rt$. Choosing the right form matters: for a fixed current use $I^{2}R$, but for a fixed voltage use $V^{2}/R$, which reverses how series versus parallel resistors share the heat.
This is why a heater's element has high resistance while connecting wires have low resistance, and why bulbs are rated by power at a stated voltage. A subtle NEET point: at constant voltage, a lower-resistance device draws more current and dissipates more power, so a 100 W bulb actually has lower resistance than a 60 W bulb of the same voltage rating.
| Law / quantity | Statement |
|---|---|
| Junction rule (KCL) | $\sum I_{in} = \sum I_{out}$ |
| Loop rule (KVL) | $\sum \Delta V = 0$ |
| Wheatstone balance | $\dfrac{P}{Q} = \dfrac{R}{S}$ |
| Power dissipated | $P = I^{2}R = V^{2}/R$ |
In a balanced Wheatstone bridge, $P = 10\ \Omega$, $Q = 20\ \Omega$, $R = 15\ \Omega$. Find the unknown $S$.
Show solution
At balance $\dfrac{P}{Q} = \dfrac{R}{S}$, so $S = \dfrac{QR}{P} = \dfrac{20\times15}{10} = 30\ \Omega$.
A heater draws $5\ \text{A}$ at $220\ \text{V}$. Find the power consumed.
Show solution
$P = VI = 220\times5 = 1100\ \text{W} = 1.1\ \text{kW}$.
Kirchhoff's junction rule is based on conservation of:
Kirchhoff's loop rule is based on conservation of:
A Wheatstone bridge is balanced when:
The power dissipated in a resistor is:
At the same voltage, a 100 W bulb compared with a 60 W bulb has:
NEET tip: Use $I^{2}R$ for fixed current, $V^{2}/R$ for fixed voltage. At balance the Wheatstone galvanometer branch carries no current and can be ignored.
Galvanometer, Ammeter, Voltmeter and PotentiometerTopic 4
A galvanometer detects small currents, deflecting in proportion to the current through its coil. On its own it cannot measure large currents or voltages, so it is converted into the meters we need by adding the right resistor — a favourite NEET topic because the logic is precise and quantitative.
To make an ammeter (which measures current and is connected in series), a small resistance called a shunt is connected in parallel with the galvanometer. The shunt carries most of the current, so only a small known fraction passes through the coil. An ideal ammeter has zero resistance so it does not disturb the circuit current.
To make a voltmeter (which measures potential difference and is connected in parallel), a large resistance is connected in series with the galvanometer. This keeps the current drawn tiny, so the voltmeter barely loads the circuit. An ideal voltmeter has infinite resistance. Mixing up these two — shunt in parallel for an ammeter, high resistance in series for a voltmeter — is a common NEET error worth drilling.
The potentiometer is a long uniform wire that measures potential difference and EMF without drawing any current at balance, making it more accurate than a voltmeter (which always draws a little). The potential drops uniformly along the wire, so the balance length is proportional to the EMF being measured. It is used to compare two EMFs and to find a cell's internal resistance — applications NEET tests with simple ratio calculations.
| Instrument | Made from galvanometer by |
|---|---|
| Ammeter (series) | small shunt in parallel; ideal $R = 0$ |
| Voltmeter (parallel) | large resistance in series; ideal $R = \infty$ |
| Potentiometer | draws no current at balance; $V \propto$ length |
Why does an ideal ammeter have zero resistance?
Show solution
An ammeter is connected in series, so any resistance it adds would reduce the circuit current it is meant to measure. Zero resistance means it does not disturb the current.
A potentiometer balances a cell of EMF $\varepsilon_1$ at $40\ \text{cm}$ and another of EMF $\varepsilon_2$ at $60\ \text{cm}$. Find the ratio $\varepsilon_1 : \varepsilon_2$.
Show solution
EMF is proportional to balance length: $\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{40}{60} = \dfrac{2}{3}$.
A galvanometer is converted into an ammeter by connecting a:
An ideal voltmeter has resistance:
An ammeter is always connected in:
The potentiometer is more accurate than a voltmeter because at balance it:
In a potentiometer, the EMF measured is proportional to the:
NEET tip: Ammeter = shunt in parallel (ideal $R = 0$, in series); Voltmeter = high resistance in series (ideal $R = \infty$, in parallel). Potentiometer draws no current at balance, so EMF $\propto$ length.
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