NCERT Exemplar · Class 10 Science Electricity

Ohm's law: at constant temperature, the current through a conductor is directly proportional to the potential difference across it. \[V\propto I \quad (T\ \text{const.}) \] \[\Rightarrow V=IR,\qquad R=\text{const.} \]Verification: Vary the current with the rheostat; note \(\displaystyle V\) (voltmeter) and \(\displaystyle I\) (ammeter) at four or five settings. \[\frac{V_1}{I_1}=\frac{V_2}{I_2}=\frac{V_3}{I_3}=\dots=R \] The points lie on a straight line through the origin; its slope is \(\displaystyle R\).Keep each reading brief — a prolonged current heats the wire and changes \(\displaystyle R\), bending the line.Not universal: it holds only at constant temperature. Diodes, electrolytes and transistors give a curved \(\displaystyle V\)-\(\displaystyle I\) graph (non-ohmic).Answer: \(\displaystyle V=IR\) at constant \(\displaystyle T\); the \(\displaystyle V\)-\(\displaystyle I\) graph is a straight line through the origin with slope \(\displaystyle R\); non-ohmic devices do not obey it.

Resistivity \(\displaystyle \rho\) of a material is the resistance of a piece of it with unit length and unit cross-sectional area, at a given temperature — a property of the material, not of a particular wire's size. \[R=\rho\frac{l}{A} \quad\Rightarrow\quad \rho=\frac{RA}{l} \] \[[\rho]=\frac{\Omega\cdot \mathrm{m}^2}{\mathrm{m}}=\Omega\,\mathrm{m} \]Experiment: measure \(\displaystyle R=V/I\) with the ammeter-voltmeter circuit, swapping the test wire each time: Pass current only briefly between trials — a warmed-up wire's \(\displaystyle R\) drifts and confounds the comparison.Answer: \(\displaystyle \rho=RA/l\), unit \(\displaystyle \Omega\,\mathrm{m}\); with the test wire swapped in the same ammeter-voltmeter circuit, \(\displaystyle R\) is found directly proportional to length, inversely proportional to area, and dependent on the material.

Connect three resistors \(\displaystyle R_1,R_2,R_3\) in series with a battery and a key. Insert an ammeter in turn at three positions — before \(\displaystyle R_1\), between \(\displaystyle R_1\) and \(\displaystyle R_2\), between \(\displaystyle R_2\) and \(\displaystyle R_3\) — and note each reading.A series loop is a single path, so charge cannot pile up anywhere along it (conservation of charge): \[I_1=I_2=I_3=I \]Use the same ammeter moved between the three positions (not three different ones), so the comparison isn't skewed by different meter resistances.The three readings come out equal, confirming the same current flows through every part.Answer: Inserting one ammeter in turn before \(\displaystyle R_1\), between \(\displaystyle R_1\)-\(\displaystyle R_2\), and between \(\displaystyle R_2\)-\(\displaystyle R_3\) gives the same reading each time — \(\displaystyle I_1=I_2=I_3\) — because a series circuit has only one path.

Connect three resistors \(\displaystyle R_1,R_2,R_3\) in parallel across a battery and key. Connect a voltmeter across each resistor in turn, then across the battery, and note the readings.All three share the same two end-points, so the potential difference across each is the same: \[V_1=V_2=V_3=V \]Put the voltmeter across the two ends of the resistor under test, never in series with it: in series it blocks that branch's current and reads the p.d. between the two junctions, not the drop across the resistor.Answer: The voltmeter reads the same value across each resistor, \(\displaystyle V_1=V_2=V_3\), equal to the reading across the battery.

When current \(\displaystyle I\) flows through resistance \(\displaystyle R\) for time \(\displaystyle t\), the electrical work done is dissipated entirely as heat: \[W=VIt=(IR)It=I^2Rt \] \[\Rightarrow H=I^2Rt \quad \text{(Joule's law of heating)} \]Demonstration: pass a measured current \(\displaystyle I\) through a resistance coil immersed in a known mass \(\displaystyle m\) of water (specific heat \(\displaystyle s\)) inside a calorimeter, for time \(\displaystyle t\); note the temperature rise \(\displaystyle \Delta T\). \[\text{Heat gained by water}=ms\Delta T \approx I^2Rt \] — the two sides matching (within losses) demonstrates the law.Lag/insulate the calorimeter and stir the water, or heat escapes before it is measured.Applications: electric heater/iron (coil's heat used directly), an incandescent bulb's filament (heated to glowing), a fuse wire (melts and breaks the circuit on overcurrent — safety), an electric kettle/geyser (heating water).Answer: \(\displaystyle H=I^2Rt\); demonstrated calorimetrically (\(\displaystyle ms\Delta T\approx I^2Rt\)); used in heaters/irons, bulb filaments, fuse wires, and kettles/geysers.