Electrical Circuits

Complete Textbook & Self-Study Guide for Grade XII NEB Board Examinations. Detailed theoretical derivations, schematic vector diagrams, worked numericals, and conceptual QA.

14.1

Kirchhoff’s Laws and Circuit Analysis

Simple electrical circuits can be analyzed using Ohm's law. However, for complex networks containing multiple loops, junctions, and EMF sources, Ohm's law alone is insufficient. In 1845, German physicist Gustav Kirchhoff formulated two fundamental laws that allow systematic analysis of complex electrical networks.

1. Kirchhoff's First Law (Current Law / KCL or Junction Rule)

Statement:

In any electric circuit, the algebraic sum of currents meeting at any junction (or node) is equal to zero.

$ \sum I = 0 $ $\implies$ $ \sum I_{\text{entering}} = \sum I_{\text{leaving}} $

Sign Convention: Currents entering a junction are taken as positive ($+$), while currents leaving a junction are taken as negative ($-$).

Figure 14.1: (a) Illustration of Kirchhoff's Current Law at node O. (b) Closed loop traversal for Kirchhoff's Voltage Law.

Referring to Fig 14.1(a), applying KCL at junction O gives:

$ (+I_1) + (+I_2) + (-I_3) + (-I_4) + (-I_5) = 0 $

$ I_1 + I_2 = I_3 + I_4 + I_5 $

Physical Significance of KCL:

KCL is a direct consequence of the Law of Conservation of Electric Charge. Since electric charge can neither be created nor destroyed at a junction, the total rate of charge entering a node must equal the total rate of charge leaving it.

2. Kirchhoff's Second Law (Voltage Law / KVL or Loop Rule)

Statement:

In any closed loop of an electric network, the algebraic sum of all electromotive forces (EMFs) is equal to the algebraic sum of the products of currents and their corresponding resistances.

$ \sum E = \sum I R $ $\iff$ $ \sum \Delta V = 0 $

Sign Conventions for Traversing Closed Loops:

Potential Drop across Resistors ($IR$): When traversing a resistor in the direction of current, potential decreases, so the product $IR$ is taken as negative ($-IR$). If traversed opposite to current, it is taken as positive ($+IR$).
Electromotive Force ($E$): When passing through a cell from its negative to positive terminal (potential rise), EMF is taken as positive ($+E$). If passing from positive to negative terminal (potential drop), EMF is taken as negative ($-E$).

Physical Significance of KVL:

KVL is a direct consequence of the Law of Conservation of Energy. Since electrostatic forces are conservative, the net work done in moving a unit positive charge around any closed path in a circuit must be zero.

Worked Example Applying KCL & KVL to Solve Circuit Parameters

Problem: Consider two cells of EMFs $E_1 = 6\text{ V}$ and $E_2 = 3\text{ V}$ with internal resistances $r_1 = 1\,\Omega$ and $r_2 = 1\,\Omega$ connected in parallel across an external resistor $R = 4\,\Omega$. Determine the currents flowing through each cell and the external resistor.

Step 1: Assign loop currents using KCL. Let current $I_1$ flow from cell $E_1$ and current $I_2$ flow from cell $E_2$. Total current through external resistor $R$ is $I = I_1 + I_2$.

Step 2: Apply KVL to Loop 1 (containing $E_1$ and $R$):

$ E_1 - I_1 r_1 - (I_1 + I_2) R = 0 $
$ 6 - I_1(1) - 4(I_1 + I_2) = 0 \implies 5I_1 + 4I_2 = 6 $ --- (Equation 1)

Step 3: Apply KVL to Loop 2 (containing $E_2$ and $R$):

$ E_2 - I_2 r_2 - (I_1 + I_2) R = 0 $
$ 3 - I_2(1) - 4(I_1 + I_2) = 0 \implies 4I_1 + 5I_2 = 3 $ --- (Equation 2)

Step 4: Solve equations simultaneously:

Multiplying Eq (1) by 5 and Eq (2) by 4:

$ 25I_1 + 20I_2 = 30 $
$ 16I_1 + 20I_2 = 12 $
Subtracting gives: $ 9I_1 = 18 \implies \mathbf{I_1 = 2\text{ A}} $

Substituting $I_1 = 2\text{ A}$ into Eq (1):

$ 5(2) + 4I_2 = 6 \implies 4I_2 = -4 \implies \mathbf{I_2 = -1\text{ A}} $

Conclusion: Current from $E_1$ is $2\text{ A}$, cell $E_2$ is actually being charged with current $1\text{ A}$ (indicated by negative sign), and total current through $R$ is $I = 2 + (-1) = \mathbf{1\text{ A}}$.

14.2

Wheatstone Bridge Circuit

The Wheatstone bridge is an accurate electrical arrangement designed by Sir Charles Wheatstone for measuring an unknown electrical resistance by comparison with known standard resistances.

Circuit Construction & Working Principle

It consists of four resistances $P$, $Q$, $R$, and $S$ connected along the four arms of a quadrilateral network ABCD. A sensitive Galvanometer $G$ with plug key $K_g$ is connected across diagonal BD, while a cell $E$ with key $K$ is connected across diagonal AC.

Figure 14.2: Wheatstone Bridge Circuit Network with Galvanometer and Battery connections.

Derivation of Balanced Bridge Condition

Let main current $I$ from battery split at junction A into $I_1$ along arm AB and $I_2$ along arm AD. At junction B, current $I_g$ passes through Galvanometer branch BD. Therefore, current along arm BC becomes $(I_1 - I_g)$ and along arm DC becomes $(I_2 + I_g)$.

Applying KVL to closed loop ABDA:

$ -I_1 P - I_g G + I_2 R = 0 $ $\implies$ $ I_1 P + I_g G = I_2 R $ --- (Equation 1)

Applying KVL to closed loop BCDB:

$ -(I_1 - I_g) Q + (I_2 + I_g) S + I_g G = 0 $ $\implies$ $ (I_1 - I_g) Q - I_g G = (I_2 + I_g) S $ --- (Equation 2)

Balanced Condition (Null Deflection):

The bridge is said to be balanced when potentials at points B and D are equal ($V_B = V_D$). Under this state, no current flows through Galvanometer ($\mathbf{I_g = 0}$).

Substituting $I_g = 0$ into Equations (1) and (2):

From Eq (1): $ I_1 P = I_2 R $ --- (Equation 3)
From Eq (2): $ I_1 Q = I_2 S $ --- (Equation 4)

Dividing Equation (3) by Equation (4):

$ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \mathbf{\frac{P}{Q} = \frac{R}{S}} $

This is the fundamental principle of the Wheatstone bridge. If three resistances are known, the unknown resistance can be accurately determined.

Condition for Maximum Sensitivity

A Wheatstone bridge is most sensitive (produces maximum deflection in galvanometer for small changes in resistance) when all four resistances P, Q, R, and S are of the same order of magnitude.

Practical Real-World Significance

Forms the operating foundation for laboratory instruments such as Meter Bridge and Post Office Box, as well as industrial sensor bridges (strain gauges and thermistors).

14.3

Meter Bridge (Slide Wire Bridge)

The Meter Bridge is the practical laboratory form of Wheatstone's bridge. It is used to measure an unknown resistance of a wire precisely and to verify the laws of series and parallel resistance combinations.

Construction & Setup

It consists of a 1-meter long uniform wire AB (made of constantan or manganin, high resistivity and low temperature coefficient) stretched tightly over a meter scale on a wooden board. Three thick L-shaped copper strips are fixed on the board creating two gaps:

Left Gap: Connected to unknown resistance $X$.
Right Gap: Connected to a standard Resistance Box $R$.
Central Strip: Connected to a sensitive Galvanometer $G$ and a sliding contact Jockey $J$.

Figure 14.3: Meter Bridge schematic setup for measuring unknown resistance X.

Derivation of Unknown Resistance Formula

Let $r$ be the resistance per unit length ($\text{cm}$) of wire AB. When the jockey is slid to point D such that Galvanometer shows null deflection ($I_g = 0$):

Resistance of wire segment AD ($l\text{ cm}$) = $P = r \cdot l$
Resistance of wire segment DB ($(100 - l)\text{ cm}$) = $Q = r \cdot (100 - l)$

According to Wheatstone's bridge principle:

$ \frac{X}{R} = \frac{P}{Q} = \frac{r \cdot l}{r \cdot (100 - l)} $

$ \mathbf{X = R \left( \frac{l}{100 - l} \right)} $

Sources of Error & End Corrections

The resistance of thick copper strips at ends A and B is not strictly zero. These are called End Resistances ($\alpha$ at end A, $\beta$ at end B in equivalent wire lengths).

Corrected Meter Bridge Formula:

$ X = R \left( \frac{l + \alpha}{100 - l + \beta} \right) $

Precaution: To minimize errors, adjust resistance $R$ such that balancing length $l$ lies near the middle of the wire ($40\text{ cm}$ to $60\text{ cm}$).

14.4

Potentiometer

A potentiometer is a high-precision instrument used to measure electromotive force (EMF) of a cell, compare EMFs of two cells, and determine the internal resistance of a primary cell without drawing any current from the circuit.

Working Principle of Potentiometer

When a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

$ V \propto l \implies \mathbf{V = k \cdot l} $

Potential Gradient ($k$): Fall of potential per unit length of wire: $ k = \frac{V}{l} = \frac{I \rho}{A} $, where $\rho$ is resistivity and $A$ is cross-sectional area.

Why Potentiometer is Preferred over an Ideal Voltmeter?

A voltmeter requires a finite current to produce deflection, thereby measuring terminal potential difference rather than actual EMF. Conversely, a potentiometer operates on the null deflection method where no current is drawn from the measured cell at balance point ($I = 0$). Thus, it acts as an ideal voltmeter with infinite effective resistance.

1. Comparison of EMFs of Two Primary Cells

Figure 14.4: Potentiometer setup for comparing EMFs of two cells E₁ and E₂.

Derivation: Connect terminal 1 and 3 (cell $E_1$ in circuit). Slide jockey to obtain balance point $J_1$ at length $l_1$:

$ E_1 = k \cdot l_1 $ --- (Equation 1)

Next, connect terminal 2 and 3 (cell $E_2$ in circuit). Slide jockey to obtain balance point $J_2$ at length $l_2$:

$ E_2 = k \cdot l_2 $ --- (Equation 2)

Dividing Eq (1) by Eq (2):

$ \mathbf{\frac{E_1}{E_2} = \frac{l_1}{l_2}} $

2. Determination of Internal Resistance ($r$) of a Primary Cell

Figure 14.5: Potentiometer setup for determining cell internal resistance r.

Step 1: Keep key $K_2$ open (open circuit). Balance length $l_1$ corresponds to cell EMF $E'$:

$ E' = k \cdot l_1 $

Step 2: Close key $K_2$ (closed circuit with shunt resistance $R$). Balance length $l_2$ corresponds to terminal voltage $V$:

$ V = k \cdot l_2 $

Ratio of EMF to terminal voltage:

$ \frac{E'}{V} = \frac{l_1}{l_2} $

Since internal resistance $r = \left( \frac{E' - V}{V} \right) R = \left( \frac{E'}{V} - 1 \right) R$:

$ \mathbf{r = R \left( \frac{l_1 - l_2}{l_2} \right)} $

14.5

Superconductors

Superconductivity is a phenomenon of quantum physics where certain materials exhibit exactly zero electrical resistance and complete expulsion of magnetic flux fields when cooled below a characteristic critical temperature.

Discovery & Critical Temperature ($T_c$)

Discovered in 1911 by Heike Kamerlingh Onnes while cooling Mercury (Hg) in liquid helium to 4.2 K. Below $T_c$, electrical resistivity abruptly drops to strictly zero ($\rho = 0$).

Critical Magnetic Field ($H_c$)

Superconductivity vanishes if an external magnetic field exceeds a critical value $H_c(T) = H_0\left[1 - (T/T_c)^2\right]$, restoring normal electrical resistance.

The Meissner Effect

When a superconducting material is cooled below its critical temperature $T_c$ inside an external magnetic field, it completely expels all magnetic flux lines from its interior.

Perfect Diamagnetism Condition:

Inside a superconductor ($T < T_c$): $\mathbf{B = 0}$ $\iff$ $ \mu_r = 0 $, Magnetic Susceptibility $ \chi = -1 $.

Figure 14.6: Expulsion of magnetic flux lines in a superconductor (Meissner Effect).

Key Applications of Superconductors

Magnetic Resonance Imaging (MRI) scanners
Maglev trains (Magnetic Levitation)
SQUIDs (Sensitive Magnetometers)
Zero-loss power transmission cables

14.6

Perfect Conductors vs. Superconductors

While both perfect conductors and superconductors exhibit zero electrical resistance ($R = 0$), they are fundamentally different thermodynamic states distinguished by their magnetic behavior.

Theoretical Difference: In a perfect conductor, Maxwell's equation gives $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$. Since $\mathbf{E} = 0$, it implies $\frac{\partial \mathbf{B}}{\partial t} = 0$, meaning magnetic flux inside is frozen/trapped. In contrast, a true superconductor actively expels magnetic flux, enforcing $\mathbf{B = 0}$ regardless of magnetic history.

Property / Feature	Perfect Conductor	Superconductor
Electrical Resistance ($R$)	Zero ($R = 0$)	Zero ($R = 0$)
Internal Magnetic Field ($B$)	Trapped/Constant ($\frac{dB}{dt} = 0$)	Strictly Zero ($B = 0$)
Meissner Effect	Does NOT exhibit Meissner Effect	Exhibits Meissner Effect
Magnetic State	Normal magnetic response	Perfect Diamagnet ($\chi = -1$)
Reversibility	State depends on cooling sequence path	Thermodynamically reversible transition

EXTRA

Conversion of Galvanometer into Ammeter and Voltmeter

A Moving Coil Galvanometer (MCG) is a highly sensitive instrument that measures small electric currents (order of $\mu\text{A}$). It has internal resistance $G$ and full-scale deflection current $I_g$.

1. Conversion of Galvanometer into an Ammeter

To convert a galvanometer into an ammeter capable of measuring large current $I$ ($I > I_g$), a very small resistance called a Shunt ($S$) is connected in PARALLEL with the galvanometer.

Figure 14.7: Shunt resistance S connected in parallel with Galvanometer G.

Mathematical Derivation:

Since Galvanometer and Shunt are in parallel, potential difference across both is equal:

$ V_G = V_S \implies I_g \cdot G = (I - I_g) \cdot S $

$ \mathbf{S = \frac{I_g \cdot G}{I - I_g}} $

Effective Ammeter Resistance ($R_A$): $ R_A = \frac{G \cdot S}{G + S} \approx S $ (extremely low resistance. An ideal ammeter has zero resistance).

2. Conversion of Galvanometer into a Voltmeter

To convert a galvanometer into a voltmeter capable of measuring potential difference $V$, a very high resistance called a Multiplier ($R$) is connected in SERIES with the galvanometer.

Figure 14.8: High resistance R connected in series with Galvanometer G.

Mathematical Derivation:

Total resistance of series combination = $G + R$. Applying Ohm's law across terminals A and B:

$ V = I_g (G + R) \implies \frac{V}{I_g} = G + R $

$ \mathbf{R = \frac{V}{I_g} - G} $

Effective Voltmeter Resistance ($R_V$): $ R_V = G + R \approx R $ (extremely high resistance. An ideal voltmeter has infinite resistance).

SOLVED

NEB Board Exam Solved Numericals

Numerical 1: Wheatstone Bridge Unknown Resistance

Question: Four resistances $P = 10\,\Omega$, $Q = 20\,\Omega$, $R = 15\,\Omega$, and $S = 25\,\Omega$ form a Wheatstone bridge network. Calculate the value of additional resistance that must be connected in parallel with $S$ to balance the bridge.

Given: $P = 10\,\Omega$, $Q = 20\,\Omega$, $R = 15\,\Omega$. Let $S'$ be required balanced resistance in arm 4.

Formula: Balanced bridge condition: $\frac{P}{Q} = \frac{R}{S'}$

Calculation:

$\frac{10}{20} = \frac{15}{S'} \implies \frac{1}{2} = \frac{15}{S'} \implies S' = 30\,\Omega$

Since existing resistance $S = 25\,\Omega$ is less than $30\,\Omega$, let parallel resistance needed be $x$:

$\frac{1}{S'} = \frac{1}{S} + \frac{1}{x} \implies \frac{1}{30} = \frac{1}{25} + \frac{1}{x} \implies \frac{1}{x} = \frac{1}{30} - \frac{1}{25} = \frac{5 - 6}{150} = -\frac{1}{150}$

Result: To increase resistance to $30\,\Omega$, a $5\,\Omega$ resistor should be added in SERIES, or if $S=40\,\Omega$, $x = 120\,\Omega$ in parallel.

Numerical 2: Meter Bridge with End Corrections

Question: In a meter bridge experiment, null point is found at distance $l = 40.0\text{ cm}$ when standard resistance box $R = 12\,\Omega$ is used. If end corrections at ends A and B are $\alpha = 0.2\text{ cm}$ and $\beta = 0.3\text{ cm}$ respectively, calculate the exact unknown resistance $X$.

Given: $R = 12\,\Omega$, $l = 40.0\text{ cm}$, $\alpha = 0.2\text{ cm}$, $\beta = 0.3\text{ cm}$

Formula: $X = R \left( \frac{l + \alpha}{100 - l + \beta} \right)$

Substitution:

$X = 12 \left( \frac{40.0 + 0.2}{100 - 40.0 + 0.3} \right) = 12 \left( \frac{40.2}{60.3} \right) = 12 \left( \frac{2}{3} \right) = 8.0\,\Omega$

Result: Unknown resistance X = 8.0 Ω

Numerical 3: Potentiometer Internal Resistance Determination

Question: A cell gives a balance length of $250\text{ cm}$ on a potentiometer wire when in open circuit. When a resistance box of $10\,\Omega$ is shunted across the cell, balance length shifts to $200\text{ cm}$. Calculate internal resistance $r$ of the cell.

Given: $l_1 = 250\text{ cm}$, $l_2 = 200\text{ cm}$, $R = 10\,\Omega$

Formula: $r = R \left( \frac{l_1 - l_2}{l_2} \right)$

Substitution:

$r = 10 \left( \frac{250 - 200}{200} \right) = 10 \left( \frac{50}{200} \right) = 10 \times 0.25 = 2.5\,\Omega$

Result: Internal resistance of cell r = 2.5 Ω

Numerical 4: Galvanometer Conversion to Ammeter and Voltmeter

Question: A galvanometer has a resistance of $G = 50\,\Omega$ and gives full scale deflection for a current $I_g = 2\text{ mA}$ ($0.002\text{ A}$). How can it be converted into: (a) An ammeter reading up to $5\text{ A}$? (b) A voltmeter reading up to $10\text{ V}$?

Part (a) Ammeter Conversion:

$S = \frac{I_g \cdot G}{I - I_g} = \frac{0.002 \times 50}{5 - 0.002} = \frac{0.1}{4.998} \approx \mathbf{0.020\,\Omega}$

Ans: Connect a shunt resistance of 0.020 Ω in PARALLEL with Galvanometer.

Part (b) Voltmeter Conversion:

$R = \frac{V}{I_g} - G = \frac{10}{0.002} - 50 = 5000 - 50 = \mathbf{4950\,\Omega}$

Ans: Connect a multiplier resistance of 4950 Ω in SERIES with Galvanometer.

CONCEPTUAL

Short Answer Conceptual Q&A

Q1: Why is Kirchhoff's Voltage Law (KVL) considered a statement of energy conservation?

Because electrostatic force is conservative, the work done in moving a charge $q$ around any closed electrical loop must be zero ($\oint \mathbf{E} \cdot d\mathbf{l} = 0$). Since potential difference represents work done per unit charge, the total energy supplied by EMF sources equals total electrical energy dissipated across resistors, validating conservation of energy.

Q2: Why is a slide wire bridge called a "Meter Bridge"?

It is called a Meter Bridge because it utilizes a wire of exactly one meter length ($100\text{ cm}$) stretched over a meter scale on a wooden board to measure unknown resistance using Wheatstone's principle.

Q3: Why is a potentiometer preferred over a voltmeter for measuring cell EMF?

A voltmeter draws a small current from the cell, thus measuring terminal potential difference ($V = E - Ir$) rather than true EMF. A potentiometer operates on null deflection where no current flows through the test cell ($I = 0$), effectively behaving as an ideal voltmeter with infinite resistance.

Q4: What is Meissner Effect and how does it distinguish superconductors from perfect conductors?

Meissner effect is the complete expulsion of magnetic flux ($B = 0$) from inside a superconductor when cooled below $T_c$. In a perfect conductor ($R = 0$), Faraday's law dictates $\frac{dB}{dt} = 0$, meaning magnetic flux becomes trapped/frozen, but not expelled. Thus, Meissner effect proves superconductivity is a unique thermodynamic state.

Q5: Why is an ammeter connected in series while a voltmeter is connected in parallel?

An ammeter measures total line current passing through a branch, so it must be connected in series and have near-zero resistance ($R_A \approx 0$) to avoid changing circuit resistance. A voltmeter measures potential difference between two nodes, so it must be connected in parallel and have high resistance ($R_V \approx \infty$) to prevent diverting current from the main path.

PRACTICE

Self-Assessment Practice Problems

1. Two resistors $10\,\Omega$ and $20\,\Omega$ are connected in parallel gaps of a meter bridge. Find the position of null point from zero end of $100\text{ cm}$ wire.

Ans: l = 33.33 cm

2. A potentiometer wire of length $4\text{ m}$ has resistance $8\,\Omega$. A cell of EMF $2\text{ V}$ and internal resistance $2\,\Omega$ is connected across it. Calculate potential gradient along wire.

Ans: k = 0.4 V/m (0.004 V/cm)

3. A galvanometer of resistance $100\,\Omega$ gives full scale deflection for $1\text{ mA}$. Find shunt resistance required to convert it into an ammeter of range $0 - 10\text{ A}$.

Ans: S = 0.01001 Ω

4. Find the value of series resistance required to convert the same galvanometer ($G = 100\,\Omega$, $I_g = 1\text{ mA}$) into a voltmeter of range $0 - 50\text{ V}$.

Ans: R = 49,900 Ω

5. Critical magnetic field of Mercury at $0\text{ K}$ is $0.041\text{ T}$ with critical temperature $T_c = 4.2\text{ K}$. Calculate critical magnetic field at $T = 2.1\text{ K}$.

Ans: Hc(2.1K) = 0.03075 T

Unit 14: Electrical Circuits