Q1. A wire of resistivity 1.7×10⁻⁸ Ω·m, length 2 m and cross-section 1×10⁻⁶ m² is used. Find its resistance.
R = ρL/A
R = (1.7×10⁻⁸ × 2) / (1×10⁻⁶)
R = 3.4×10⁻⁸ / 10⁻⁶ = 0.034 Ω
The chapter that powers half the paper. Use the calculator to feel V = IR and electrical power, then drill EMF and resistivity numericals.
Set the voltage and resistance to read the current and power dissipated.
Power lost as heat in a wire is I²R — that is why transmission uses high voltage (low current) to cut losses.
Ohm's law, power and EMF problems with worked steps.
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Q1. A wire of resistivity 1.7×10⁻⁸ Ω·m, length 2 m and cross-section 1×10⁻⁶ m² is used. Find its resistance.
R = ρL/A
R = (1.7×10⁻⁸ × 2) / (1×10⁻⁶)
R = 3.4×10⁻⁸ / 10⁻⁶ = 0.034 Ω
Q2. A battery of EMF 10 V and internal resistance 0.5 Ω drives a current of 2 A. Find the terminal voltage.
V = E − Ir
V = 10 − (2 × 0.5)
V = 10 − 1 = 9 V
| Quantity | Formula | SI unit |
|---|---|---|
| Drift current | I = n A e v_d | ampere (A) |
| Ohm's law | V = I R | — |
| Resistivity | R = ρ L / A | ohm (Ω) |
| EMF & internal resistance | E = I (R + r) | volt (V) |
| Electric power | P = VI = I²R = V²/R | watt (W) |
| Wheatstone balance | P/Q = R/S | — |
The small average velocity (~10⁻⁴ m/s) with which free electrons move through a conductor under an applied electric field, giving I = nAev_d.
EMF is the maximum potential difference of a cell with no current flowing; terminal voltage is the actual voltage across its terminals when current flows, V = E − Ir.
Higher temperature means more vigorous lattice vibrations, which scatter the drifting electrons more, raising the resistivity and hence the resistance.
The junction rule (sum of currents at a junction is zero — charge conservation) and the loop rule (sum of potential differences around a loop is zero — energy conservation).
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