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NEET UG · Physics

Current electricity

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NEET UG · Physics

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Check cell emf, terminal voltage, branch currents and circuit power with a consistent model. Current electricity: include the loaded cell before the shortcut

About 6 min · NEET UG

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Current electricity: include the loaded cell before the shortcut

Check cell emf, terminal voltage, branch currents and circuit power with a consistent model.

What you will understand

For NEET UG Physics students: distinguish cell emf from loaded terminal voltage, balance loop energy and branch currents, and interpret drift speed without confusing it with circuit response.

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Quick answer: include the whole circuit before treating a voltage as fixed

A discharging real-cell model includes an emf in series with internal resistance. The terminal voltage across the external circuit is emf minus the internal current-dependent drop. A changed load can therefore change both total current and terminal voltage. Junction rules account for charge in steady state; loop rules account for energy per charge. Electron drift is an average carrier motion, while electrical response involves the field throughout the circuit.

Loaded cell: emf and terminal voltage label different parts

Original closed-loop steady-DC circuit with an ideal12 V emf and 1 ohm internal resistance in series with a5 ohm external load. The long source plate is the positive side feeding the internal resistor and terminal A, then the external load to terminal B and the return. Conventional current is clockwise,2 A. Terminal voltage VA-VB across the load is 12-2(1)=10 V, also2(5)=10 V. An ideal voltmeter across A/B reads10 V rather than the 12 V emf while the cell is discharging under this load. Circuit layout is schematic and the resistances are supplied constant values, not measured battery parameters or a practical wiring instruction.
The ideal 12 V emf and 1 Ω internal resistance are in series with the external 5 Ω load. Conventional current is clockwise and equals 2 A. Terminal A is 10 V above B under this load, so an ideal voltmeter across them reads 10 V. Supplied component values and ideal wires define this model.

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Find current, terminal voltage and the internal power loss

With emf ε=12.0 V, internal r=1.00 Ω and external R=5.00 Ω, total resistance is 6.00 Ω. Thus I=2.00 A and terminal V=ε−Ir=10.0 V, also IR=10.0 V. Load power is I²R=20.0 W, internal dissipation is I²r=4.00 W and ideal-emf power is εI=24.0 W. The power sum closes. If R becomes 11.0 Ω, I falls to 1.00 A while terminal voltage rises to 11.0 V; higher terminal voltage in this comparison does not mean greater current or greater load power.

Open-circuit voltage and discharging voltage are not the same comparison

In the ideal series-emf/internal-r model, an open circuit has I=0 and terminal voltage equal to ε. A load draws current and lowers the discharging terminal voltage by Ir. A nonideal voltmeter itself draws some current, so the word ideal matters when treating its reading as nonloading. Charging uses a different direction of current and the appropriate sign convention; do not carry the discharging subtraction into every battery situation without checking the loop.

A loop check tracks potential changes and power consistently

Original potential-step map around the supplied steady loop, following conventional current. Starting at chosen0 V reference, the ideal emf raises potential by 12 V, the 1 ohm internal resistor drops2 V at 2 A and the 5 ohm load drops10 V, returning to zero. Horizontal position labels circuit sequence, not time or distance, and abrupt steps represent lumped components rather than microscopic fields. Per coulomb the source provides12 J,2 J becomes internal-resistor energy and 10 J reaches the load. At 2 C/s these correspond to24 W,4 W and 20 W. Both signed loop sum and power balance close. An absolute potential zero is conventional.
Following current from the chosen zero reference, the ideal emf gives +12 V, internal resistance gives −2 V and the external load gives −10 V. The sum is zero. Per coulomb, the corresponding transfers are +12 J,−2 J,−10 J. At 2 C/s these become24 W,4 W and 20 W. Horizontal position in the diagram is loop sequence rather than time or physical distance.

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Add a parallel branch: total current can rise while each branch current falls

Use the same ε=12.0 V and r=1.00 Ω cell. Two external10.0 Ω resistors in parallel have equivalent R=5.00 Ω, so total current is 2.00 A, terminal voltage 10.0 V and each branch current 1.00 A. Add a third identical parallel resistor. Now R=10/3 Ω and I=12/(1+10/3)=36/13 AI=12/(1+10/3)=36/13\,\mathrm A. Terminal voltage is 120/13 V≈9.23 V120/13\,\mathrm V\approx9.23\,\mathrm V, so each branch current is 12/13 A≈0.923 A12/13\,\mathrm A\approx0.923\,\mathrm A. Three branch currents sum to 36/13 A. The source current increases while each identical branch current decreases, because internal resistance prevents terminal voltage from staying fixed.

Use junction and loop rules with explicit directions

Choose branch-current directions before writing the equations. In steady state, total conventional current entering a junction equals that leaving it; this does not require all branch currents to be equal. Across a resistor along its conventional current, potential falls by IR. Across the ideal emf from its negative to positive terminal, potential rises by ε. A negative solved current means that its actual direction is opposite the chosen one, not that the circuit is impossible.

A small drift speed can carry a substantial current

For a supplied metal-wire model use carrier density n=1.00×1029 m−3n=1.00\times10^{29}\,\mathrm{m^{-3}}, cross-section

A=1.00 mm2=1.00×10−6 m2A=1.00\,\mathrm{mm^2}=1.00\times10^{-6}\,\mathrm{m^2}
, I=2.00 A and electron-charge magnitude e=1.60×10−19 Ce=1.60\times10^{-19}\,\mathrm C. From I=neAvdI=neAv_d, drift-speed magnitude is
vd=1.25×10−4 m s−1=0.125 mm s−1v_d=1.25\times10^{-4}\,\mathrm{m\,s^{-1}}=0.125\,\mathrm{mm\,s^{-1}}
. Electron drift points opposite conventional current. This average is not the speed at which a change of electric field propagates through the circuit, and it is not every electron’s instantaneous microscopic speed.

Resistance ratios need the quantities being held fixed

For a uniform ohmic wire in the stated conditions, R=ρL/A. Stretching a wire can change L and A together; changing temperature can change ρ. The simple fixed-ρ relation should not be extended silently across heating or material changes. Similarly, two parallel branches share the same endpoint voltage, while series components carry the same steady current. Which voltage stays constant depends on whether the source is ideal or has the internal resistance being modelled here.

Common mistakes: a remembered 12 V label is not always the load voltage

Include internal resistance before finding current. Distinguish emf, terminal voltage and a resistor’s drop. Check a circuit by both signed potential changes and power. Parallel branches share terminal voltage, but that voltage can change when a real-cell load changes. Do not identify slow electron drift with the full electrical response or set every branch current equal without using the resistances.

Try it: increase the external resistance

For the supplied12 V cell with 1 Ω internal resistance, replace the5 Ω external load by 11 Ω. Find current, terminal voltage and external power. Does the increased terminal voltage contradict a lower current?

Reveal the answer

Current is 12/(11+1)=1 A, terminal voltage is 12−1(1)=11 V and external power is 1²(11)=11 W. Compared with the original2 A,10 V,20 W case, terminal voltage rises while current and load power fall. The internal voltage drop became smaller; there is no contradiction.

Try it: a negative solved branch current

You assumed a branch current of positive direction A to B, but solving consistent circuit equations gives −0.30 A. What does the result mean, and should you discard the solution solely because it is negative?

Reveal the answer

The actual conventional current is 0.30 A from B to A, opposite the assumed direction. Keep the sign convention consistently in the equations and use the result to interpret the direction. A negative value alone is not evidence that charge conservation or the circuit solution failed.

Come back tomorrow: check the loaded source and the energy sum

Explain why terminal voltage differs from emf, then add one parallel branch and predict which currents change. Revisit NEET Electric Potential and Capacitance for energy per charge, or NEET Semiconductor Devices to test a circuit’s device assumptions.

Chapter practice opens the broad NEET Current Electricity pool.

Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, worked examples and self-checks are original and use stated models and supplied inputs. Human subject review is not claimed. The existing chapter catalogue and these focused guides do not establish complete current exam or syllabus coverage.

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