Conservation at nodes and loops, then one careful initial condition for every capacitor.
JEE Main · JEE Advanced · NEET UG75 min study guide9 indexed sections
BY THE END, YOU CAN
Analyse source terminal voltage, internal dissipation and simple battery combinations.
Solve resistor networks with node or loop equations and justified bridge balance.
Find a Thévenin equivalent and distinguish maximum power from efficiency.
Derive and apply RC charging, discharging, switching and energy relations.
BEFORE YOU START
Ohm’s law, resistor combinations and electrical power.
Capacitance, stored field energy, simple exponentials and logarithms.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: NEET UG 2023. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Circuit Analysis.
NEET UG · 2023 · 7 May · Booklet E1 · Physics · Q7
Paraphrased exam question
Answer checked against the official final key. Worked solution written for this lesson.
A 10 V ideal source feeds 400 Ω and R in series. Their junction connects through a galvanometer to the positive terminal of a 2 V ideal source; both negative terminals share the lower node. At zero galvanometer current, find R.
Reveal answer & worked solution
Answer: R = 100 Ω (original option 3).
Zero galvanometer current means no voltage drop across it: the junction is 2 V above the common lower node.
The 400 Ω resistor therefore drops 10 − 2 = 8 V, giving I = 8/400 = 0.020 A.
No current leaves through the galvanometer, so that same current enters R. Hence R = 2/0.020 = 100 Ω.
Source: NEET (UG) 2023, 7 May, E1, Q7; official final answer key. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Replace the drawing with a map of nodes
A circuit diagram is a connectivity map, not a picture of the physical layout. Every point joined by an ideal uninterrupted wire belongs to the same node and has the same potential in the lumped-circuit approximation. Marking nodes often reveals that components drawn far apart are parallel, while adjacent-looking components are not in series because a junction lies between them.
Choose a reference node and assign it zero potential. Only potential differences affect ordinary circuit operation; this reference does not require an actual connection to Earth. Label unknown node potentials and branch currents. You may guess current directions arbitrarily: a negative solution means the actual direction is opposite to the arrow.
An ideal wire has zero resistance, an ideal ammeter zero resistance and an ideal voltmeter infinite resistance. These are modelling limits. A short circuit across a finite resistor forces its voltage to zero in an ideal circuit, while an open branch carries zero current. An open branch can still have a nonzero potential difference, as at the terminals of an unloaded battery.
Before writing equations, ask which components store energy, which dissipate it and which supply it. Resistors relate instantaneous voltage and current through V = IR; capacitors relate voltage to accumulated charge. A capacitor cannot be replaced by its long-time open-circuit behaviour immediately after a switch changes position.
02 / UNDERSTAND & APPLY
2. EMF describes source action, not always terminal voltage
Electromotive force ε is energy supplied by a source per unit charge through its non-electrostatic mechanism. Despite the name, it is measured in volts, not newtons. A real cell is often modelled as an ideal emf in series with internal resistance r. Its terminal voltage depends on current and on whether it is delivering or absorbing electrical energy.
For a discharging cell with current I leaving its positive terminal, Vterminal = ε − Ir. Some chemical energy is dissipated internally, leaving less voltage for the external circuit. With a load R, I = ε/(R + r). For charging current entering the positive terminal, the applied terminal voltage is ε + Ir in the same simple model.
Open-circuit terminal voltage equals ε in this model because I = 0. A short circuit gives I = ε/r if r is finite, but a real battery may deviate strongly from the simple constant-r model and shorting it is dangerous. Treat such calculations as idealised analysis, not instructions for a physical experiment.
Cells in series add their signed emfs and their internal resistances. Identical cells connected in parallel with matching polarity keep the same emf while reducing equivalent internal resistance. Unequal ideal voltage sources cannot simply be placed in parallel without resolving internal resistance and possible circulating currents.
ORIGINAL WORKED EXAMPLE
Terminal voltage under load
A cell with ε = 12 V and r = 1 Ω supplies a 5 Ω resistor. Find current, terminal voltage and internal heating power.
Total resistance is 6 Ω, so current is 12/6 = 2 A.
Terminal voltage is ε − Ir = 12 − 2 = 10 V, matching IR = 2 × 5.
Internal heating is I²r = 4 W. The external load receives I²R = 20 W and source chemical power is εI = 24 W.
Answer 2 A; 10 V; 4 W internal heating.
03 / UNDERSTAND & APPLY
3. Kirchhoff’s rules encode conservation
The junction rule follows charge conservation: for a node without accumulating charge, total current entering equals total current leaving. Assign signs consistently rather than deciding by visual arrow count. The loop rule follows energy conservation in the electrostatic lumped-circuit model: the algebraic sum of potential changes around a closed loop is zero.
Crossing a resistor in the assumed current direction contributes a potential drop −IR; crossing against it contributes +IR. Crossing an ideal source from its negative terminal to positive contributes +ε, and the reverse contributes −ε. Traversal direction is your choice and need not match the current direction. Reversing an entire loop equation produces the same physics.
Write only independent equations. With branch-current analysis, junction equations connect branch currents and independent loop equations connect voltage drops. In node-voltage analysis, each resistor current from node a to node b is (Va − Vb)/R, and the junction rule gives linear equations in the unknown node voltages. This method is often cleaner for multiple parallel branches with sources.
The simple zero-sum electrostatic loop rule needs modification when a changing magnetic flux links the circuit: induced emf must be included. The circuit methods here assume a lumped network where that effect is either negligible or already represented by an explicit source or inductor model. State this boundary rather than claiming every real loop always has zero electric circulation.
ΣIin = ΣIout; ΣΔVloop = 0
Steady or lumped-node charge balance and energy balance with all sources included.
Iab = (Va − Vb)/R
Current through a resistor from node a to node b for the chosen positive direction.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Terminal potentials are imposed by ideal supplies sharing the same reference. Diagram is a topology schematic.
ORIGINAL WORKED EXAMPLE
A node connected to two supplies
A node V connects through 2 Ω to a fixed +10 V node, through 3 Ω to a fixed +5 V node, and through 6 Ω to ground. Find V and branch currents.
Write outward currents: (V − 10)/2 + (V − 5)/3 + V/6 = 0.
Multiply by 6: 3V − 30 + 2V − 10 + V = 0, so V = 20/3 V.
Current from 10 V toward the node is (10 − 20/3)/2 = 5/3 A. Current from the node toward 5 V is (20/3 − 5)/3 = 5/9 A.
Current to ground is (20/3)/6 = 10/9 A. Incoming 5/3 equals outgoing 5/9 + 10/9.
Answer V = 20/3 V; currents 5/3 A in, 5/9 A to the 5 V node, 10/9 A to ground.
04 / UNDERSTAND & APPLY
4. Use symmetry and balance only when justified
In a Wheatstone-style bridge, two series branches connect the same supply nodes and a detector joins their midpoint nodes. If the ratios of the upper to lower resistances match, the midpoint potentials are equal and detector current is zero. This follows from voltage division; it is not a new conservation law.
Label upper-left R₁, lower-left R₂, upper-right R₃ and lower-right R₄ between common top and bottom supply nodes. Balance requires R₁/R₂ = R₃/R₄. At balance the detector branch can be omitted for calculating the other currents because its voltage and current are zero. Off balance it generally loads the divider and must remain in the circuit equations.
Symmetry can identify equipotential nodes when exchanging them leaves the sources and resistors unchanged. Equal-looking components alone are insufficient if the source placement breaks the symmetry. Once two nodes are proven equipotential, no current flows through a resistor directly joining them. A zero-current resistor is not the same as an absent voltage source; inspect each component’s constitutive relation.
ORIGINAL WORKED EXAMPLE
Balance a bridge
Two bridge branches run from +12 V to ground. The left has 2 kΩ above its midpoint and 4 kΩ below. The right has 3 kΩ above and an unknown R below. Find R for zero detector current and the midpoint voltage.
Balance requires 2/4 = 3/R, giving R = 6 kΩ.
Left branch current is 12 V/6 kΩ = 2 mA.
Midpoint voltage is the drop across its lower 4 kΩ: 2 mA × 4 kΩ = 8 V. The right midpoint is also 8 V.
Answer R = 6 kΩ; both midpoints at 8 V.
05 / UNDERSTAND & APPLY
5. Simplify what the capacitor or load sees
A linear resistive network with independent sources can often be replaced at two terminals by a Thévenin equivalent: an open-circuit voltage Vth in series with resistance Rth. Find Vth with the load removed. Find Rth by suppressing independent ideal sources: replace ideal voltage sources by shorts and ideal current sources by opens, retaining their modelled internal resistances.
This source-suppression shortcut needs care with dependent sources, which must remain active and may require a test source. The present examples use independent sources only. The equivalent reproduces the terminal voltage–current relation; it does not reproduce every internal current or internal dissipation of the original network.
For a load R connected to a source ε with series resistance r, load power is ε²R/(R + r)². Differentiating with respect to positive R gives maximum power at R = r. At that point half the source power dissipates in the internal resistance, so maximum power transfer is not maximum efficiency. Very large R gives high fractional efficiency but small delivered power.
ORIGINAL WORKED EXAMPLE
A loaded divider as an equivalent source
A 12 V source feeds a divider of 2 kΩ above the output and 4 kΩ below it. Find its Thévenin equivalent and output when a 4 kΩ load is attached to ground.
Open-circuit output is Vth = 12 × 4/(2 + 4) = 8 V.
Suppress the ideal supply to a short. The output sees 2 kΩ and 4 kΩ in parallel, so Rth = 4/3 kΩ.
With the 4 kΩ load, output is 8 × 4/(4 + 4/3) = 6 V. Loading changes the divider from its 8 V open-circuit value.
6. Derive the charging exponential from a loop equation
Consider an initially uncharged capacitor C connected through resistor R to an ideal constant source ε at t = 0. Take current in the charging direction and capacitor voltage VC = q/C. The loop equation is ε − IR − q/C = 0, with I = dq/dt. Rearranging gives dq/dt = (ε − q/C)/R: charging slows as the capacitor voltage grows.
Solving with q(0) = 0 gives q(t) = Cε[1 − exp(−t/RC)] and I(t) = (ε/R)exp(−t/RC). The time constant τ = RC sets the timescale. After one τ, charge reaches 1 − e⁻¹ ≈ 63.2% of its final value while current falls to e⁻¹ ≈ 36.8% of its initial value. It never reaches the exact final value at finite time in the ideal exponential model.
At the instant after connection, an initially uncharged ideal capacitor has zero voltage and behaves like a short only for determining that initial circuit state with finite resistance. At long times in a DC resistive network, its current is zero and it behaves as an open branch. These are limiting behaviours, not universal replacements throughout the transient.
For arbitrary initial voltage V₀ and a fixed Thévenin source V∞ through Rth, the capacitor voltage is VC(t) = V∞ + (V₀ − V∞)exp[−t/(RthC)]. This form makes initial and final conditions explicit. The current direction may reverse depending on whether V₀ is above or below the new final voltage.
VC(t) = V∞ + (V₀ − V∞)e^(−t/τ); τ = RthC
One-capacitor linear first-order circuit with fixed sources after switching and Rth > 0.
I(t) = (ε/R)e^(−t/RC)
Initially uncharged capacitor charging through R from an ideal constant source ε.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Initially uncharged in the charging example. After switching to B, the same R discharges C. The switch never joins A and B simultaneously.
ORIGINAL WORKED EXAMPLE
Time to a specified charging fraction
An initially uncharged 100 μF capacitor charges through 20 kΩ from an ideal 10 V source. Find the time to reach 8 V and current at that instant.
Time constant is RC = (20 × 10³)(100 × 10⁻⁶) = 2 s.
Set 8 = 10(1 − e^(−t/2)), so e^(−t/2) = 0.2 and t = 2 ln5 ≈ 3.22 s.
The resistor has 10 − 8 = 2 V across it, so current is 2/20000 = 0.10 mA.
Answer Approximately 3.22 s; current 0.10 mA.
07 / UNDERSTAND & APPLY
7. Discharge and energy accounting complete the story
When a charged capacitor with initial voltage V₀ discharges through R, there is no source maintaining the voltage. The capacitor voltage and charge decay as e^(−t/RC). Current magnitude is (V₀/R)e^(−t/RC), flowing in the discharge direction. If current was defined in the earlier charging direction, its algebraic sign is negative.
Capacitor energy is U = CV²/2. Because voltage decays exponentially, stored energy decays as e^(−2t/RC), with an energy e-folding time RC/2. The charge half-life is RC ln2, while the energy half-life is (RC/2)ln2. Confusing voltage decay with energy decay produces a factor-of-two error.
For charging an initially uncharged capacitor from an ideal fixed voltage ε through a resistor, the source supplies total energy εQfinal = Cε². Final field energy is Cε²/2, and the other half is dissipated in the resistor. This result is independent of R in the ideal complete-charging model; R changes the timescale and power profile. Letting R approach zero makes the ideal transient singular, so real parasitic inductance and radiation become relevant.
At a switching instant, an ideal capacitor’s voltage cannot jump without an impulsive current. With only finite resistances and finite sources, carry the pre-switch voltage into the post-switch circuit as the initial condition. If an ideal source is directly connected across a capacitor at a different voltage, the ideal model predicts an impulse and needs physical resistance or inductance to resolve the transient.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Quantitative curves. Energy is proportional to voltage squared; solid and dashed lines distinguish the quantities.
ORIGINAL WORKED EXAMPLE
Discharge energy versus charge
A 200 μF capacitor initially at 20 V discharges through 5 kΩ. Find remaining voltage and energy after one time constant.
Time constant is 5000 × 200 × 10⁻⁶ = 1 s.
At t = τ, voltage is 20/e ≈ 7.36 V.
Initial energy is (1/2)(200 × 10⁻⁶)(20²) = 0.040 J. Remaining energy is 0.040/e² ≈ 0.00541 J.
Answer Voltage 7.36 V; energy about 5.41 mJ.
08 / UNDERSTAND & APPLY
8. Original practice: steady state and switching
Keep initial conditions, current arrows and source signs visible. Each problem refers to ideal lumped components with any internal resistance explicitly stated.
ORIGINAL PRACTICE · 1
A 6 V cell with 0.5 Ω internal resistance drives 2.5 Ω. Find terminal voltage.
Reveal answer & reasoning
5 V.
Current is 6/3 = 2 A, so terminal voltage is 6 − 2(0.5) = 5 V.
ORIGINAL PRACTICE · 2
A cell with emf 4 V and internal resistance 1 Ω is charged at 0.5 A. What terminal voltage is required in the simple model?
Reveal answer & reasoning
4.5 V.
Charging current enters the positive terminal, so V = ε + Ir.
ORIGINAL PRACTICE · 3
At a junction, 3 A and 2 A enter while one branch carries 4 A out. Find the fourth branch current.
Reveal answer & reasoning
1 A outward.
Charge conservation requires 5 A total outward.
ORIGINAL PRACTICE · 4
A resistor has node potentials 9 V and 3 V at its ends and resistance 2 Ω. Find current direction and magnitude.
Reveal answer & reasoning
3 A from 9 V to 3 V.
I = (9 − 3)/2.
ORIGINAL PRACTICE · 5
A balanced bridge has upper/lower resistances 3 Ω/6 Ω on the left and 5 Ω/R on the right. Find R.
Reveal answer & reasoning
10 Ω.
Equal midpoint potentials require 3/6 = 5/R.
ORIGINAL PRACTICE · 6
A source ε with internal resistance r > 0 drives a variable load. At maximum load power, what is R and source-to-load efficiency?
Reveal answer & reasoning
R = r; 50%.
The two equal resistances dissipate equal power, so half reaches the load.
ORIGINAL PRACTICE · 7
A 2 μF capacitor charges through 3 MΩ. Find time constant.
Reveal answer & reasoning
6 s.
Multiply 2 × 10⁻⁶ F by 3 × 10⁶ Ω.
ORIGINAL PRACTICE · 8
An initially uncharged capacitor charges toward 12 V. What is its voltage after one time constant?
Reveal answer & reasoning
12(1 − e⁻¹) ≈ 7.59 V.
The charging fraction after τ is approximately 0.632.
ORIGINAL PRACTICE · 9
A discharging capacitor’s voltage halves. What fraction of its initial energy remains?
Reveal answer & reasoning
One quarter.
Energy is proportional to voltage squared at fixed capacitance.
ORIGINAL PRACTICE · 10
A capacitor is at 8 V just before a switch change into a finite-resistance network. What is its voltage immediately afterward?
Reveal answer & reasoning
8 V.
Finite current cannot transfer a finite charge in zero time, so capacitor voltage is continuous.
ORIGINAL PRACTICE · 11
A 10 μF capacitor charges fully from zero through a resistor using a 6 V ideal source. Find supplied, stored and dissipated energies.
Source supplies Cε²; capacitor stores half of that.
ORIGINAL PRACTICE · 12
A capacitor initially at 10 V is connected through resistance to a 4 V Thévenin source. State VC(t).
Reveal answer & reasoning
VC(t) = 4 + 6e^(−t/RC) V.
Use final plus initial-minus-final times the decaying exponential; voltage starts at 10 and tends to 4.
09 / UNDERSTAND & APPLY
9. Check conservation and every limiting case
After solving a resistive network, test junction current balance and power balance. A negative current is acceptable if its interpretation is consistent. A negative dissipation I²R for a positive resistor is not. Terminal voltage, emf and resistor drops should tell the same energy story around each independent loop.
For RC motion, test t = 0, t → ∞ and the units of the time constant. The capacitor starts at its pre-switch voltage and approaches the new network’s open-circuit terminal voltage. Resistance seen by the capacitor sets the timescale. This initial/final/evolution pattern often solves a complicated-looking switch problem more reliably than memorising separate charging and discharging formulas.
Identify nodes before reducing a network.
EMF and terminal voltage differ when internal resistance carries current.
Arbitrary current directions are allowed if signs remain consistent.
Bridge balance is a statement about equal midpoint potentials.
A finite-current capacitor voltage is continuous at switching.
Stored energy decays twice as fast exponentially as discharge voltage.
KEEP THE BIG IDEAS
Quick revision
Circuit topology decides which quantities are shared.
Current direction guesses are bookkeeping, not assumptions about the final answer.
A capacitor’s initial voltage and the new steady voltage determine its transient.
Energy accounting checks both steady circuits and charging/discharging processes.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.