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Current electricity is one of the most consistently tested NEET Physics chapters, and most of it comes down to applying Ohm's law and Kirchhoff's laws carefully in circuits with multiple resistors, cells or a Wheatstone bridge.
NEET weightage: High · NCERT Physics Class 12, Part 1, Chapter 3
For an ohmic conductor at constant temperature, V = IR, where R depends on the conductor's material and geometry: R = ρL/A (ρ is resistivity, L is length, A is cross-sectional area). Resistivity is a material property; resistance also depends on shape.
In series, the same current flows through every resistor, and the equivalent resistance is the sum: R_eq = R1 + R2 + R3 + .... Voltage divides across resistors in proportion to their resistance.
In parallel, the voltage across every resistor is the same, and the reciprocal of the equivalent resistance is the sum of reciprocals: 1/R_eq = 1/R1 + 1/R2 + 1/R3 + .... Current divides between branches inversely proportional to resistance.
Kirchhoff's Current Law (KCL): the total current entering a junction equals the total current leaving it (charge conservation). Kirchhoff's Voltage Law (KVL): the sum of potential differences around any closed loop is zero (energy conservation). These let you solve circuits too complex for simple series/parallel reduction.
A four-resistor bridge circuit used to find an unknown resistance. At balance, no current flows through the galvanometer arm, and P/Q = R/S (where P, Q, R, S are the four arm resistances). This balance condition is a frequently tested NEET formula.
A real cell has internal resistance r. When it drives current I through an external circuit, the terminal voltage is V = ε − Ir (ε is the EMF). Terminal voltage is always less than EMF while current flows, and equals EMF only when the circuit is open (I = 0).
Ohm's law
V = IR
Resistance from resistivity
R = ρL / A
Series equivalent resistance
R_eq = R1 + R2 + …
Parallel equivalent resistance
1/R_eq = 1/R1 + 1/R2 + …
Wheatstone bridge balance condition
P/Q = R/S
Terminal voltage with internal resistance
V = ε − Ir
Electrical power
P = VI = I²R = V²/R
Two resistors, 4 Ω and 6 Ω, are connected in parallel. What is the equivalent resistance?
1/R_eq = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so R_eq = 12/5 = 2.4 Ω. Notice this is less than the smaller resistor (4 Ω) — a useful sanity check for any parallel combination.
A cell of EMF 12 V and internal resistance 1 Ω is connected to an external resistor of 5 Ω. What is the current in the circuit, and the terminal voltage?
Total resistance = internal + external = 1 + 5 = 6 Ω. Current I = ε / R_total = 12/6 = 2 A. Terminal voltage V = ε − Ir = 12 − (2×1) = 10 V. Notice the terminal voltage (10 V) is less than the EMF (12 V) because current is flowing.
Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. What is the equivalent resistance?
1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R_eq = 1 Ω. As expected, it's smaller than the smallest individual resistor (2 Ω).
In a Wheatstone bridge, the four arm resistances are P = 10 Ω, Q = 20 Ω, R = 15 Ω, and S is unknown. If the bridge is balanced, what is S?
At balance, P/Q = R/S, so 10/20 = 15/S. Solving: S = 15 × 20/10 = 30 Ω.
A wire of resistivity 1.7 × 10⁻⁸ Ω·m, length 2 m and cross-sectional area 1 × 10⁻⁶ m² carries current. What is its resistance?
R = ρL/A = (1.7 × 10⁻⁸ × 2) / (1 × 10⁻⁶) = 3.4 × 10⁻⁸ / 10⁻⁶ = 0.034 Ω.
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