Circuit Analysis Techniques

Kirchhoff’S Rules ($ \Sigma I = 0 $, $ \Sigma V = 0 $)

Analysing simple circuits with resistors in series and parallel can often be done using Ohm's Law directly. However, for more complex circuits, such as those with multiple sources of EMF or where resistors are not strictly in series or parallel combinations (e.g., a Wheatstone bridge circuit), Ohm's Law alone is insufficient. For such circuits, a more general set of rules, known as Kirchhoff's Rules (or Kirchhoff's Laws), is used. These rules are based on the fundamental conservation principles of electric charge and energy. There are two Kirchhoff's Rules:

1. Kirchhoff's Junction Rule (or Current Law - KCL)
2. Kirchhoff's Loop Rule (or Voltage Law - KVL)

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Kirchhoff's Junction Rule ($ \Sigma I = 0 $)

Kirchhoff's Junction Rule states that:

"The algebraic sum of the currents entering any junction (or node) in an electric circuit is equal to the algebraic sum of the currents leaving that junction."

Alternatively, the rule can be stated as:

"The algebraic sum of all currents meeting at a junction is zero."

$ \Sigma I = 0 $ (at a junction)

Sign Convention: For applying this rule, a sign convention is used. Typically, currents entering the junction are taken as positive, and currents leaving the junction are taken as negative (or vice-versa). For example, if currents $I_1, I_2$ are entering a junction and $I_3, I_4, I_5$ are leaving, the rule states $I_1 + I_2 - I_3 - I_4 - I_5 = 0$, or $I_1 + I_2 = I_3 + I_4 + I_5$.

Kirchhoff's Junction Rule at a junction: $I_1 + I_2 = I_3 + I_4$.

Basis: The Junction Rule is a direct consequence of the conservation of electric charge. Charge cannot accumulate at any point (junction) in a circuit. The total charge entering a junction per unit time must equal the total charge leaving the junction per unit time. Since current is the rate of flow of charge, the sum of currents entering equals the sum of currents leaving.

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Kirchhoff's Loop Rule ($ \Sigma V = 0 $)

Kirchhoff's Loop Rule states that:

"The algebraic sum of the changes in potential around any closed loop (or mesh) in an electric circuit is zero."

Kirchhoff's Loop Rule applied to a closed loop.

Sign Conventions for Potential Change: To apply the Loop Rule, we traverse a closed loop and add up the potential changes across each component.

• Across a Resistor (R):
  • If traversing in the direction of the current ($I$), the potential drops. Change in potential is $-IR$.
  • If traversing against the direction of the current ($I$), the potential rises. Change in potential is $+IR$.
• Across a Cell/Source of EMF ($\mathcal{E}$):
  • If traversing from the negative terminal to the positive terminal (in the direction of EMF rise within the source), the potential rises. Change in potential is $+\mathcal{E}$.
  • If traversing from the positive terminal to the negative terminal (against the direction of EMF rise within the source), the potential drops. Change in potential is $-\mathcal{E}$.

Let's illustrate with a simple loop containing a battery and a resistor:

Simple loop with a battery ($\mathcal{E}$, internal resistance $r$) and external resistor ($R$). Current $I$ flows clockwise.

Assuming clockwise current $I$, let's traverse the loop clockwise starting from point A.

• From A to B (across external resistor R): Traverse in direction of $I$, potential drop. Change is $-IR$.
• From B to C (across internal resistance r of cell): Traverse in direction of $I$, potential drop. Change is $-Ir$.
• From C to A (across the battery from negative to positive terminal): Traverse in direction of EMF rise. Change is $+\mathcal{E}$.

Sum of potential changes around the loop A-B-C-A:

$ -IR - Ir + \mathcal{E} = 0 $

$ \mathcal{E} = IR + Ir = I(R+r) $

$ I = \frac{\mathcal{E}}{R+r} $

This is the same formula we derived earlier for the current from a cell with internal resistance connected to an external resistor, confirming the validity of the Loop Rule and sign conventions.

Basis: The Loop Rule is a direct consequence of the conservation of energy. When a charge moves around a closed loop and returns to its starting point, the net work done on it by the electric field is zero. This means the sum of potential rises must equal the sum of potential drops around any closed loop. Electric potential is a scalar quantity and is path-independent in a static electric field (which is the case in DC circuits).

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Solving Circuits using Kirchhoff's Rules

To solve a complex circuit (find currents through each branch, potential difference across components):

1. Identify all junctions and independent loops in the circuit.
2. Assign a direction to the current in each branch. If the chosen direction is wrong, the calculated current will be negative, indicating the actual direction is opposite.
3. Apply Kirchhoff's Junction Rule at each independent junction. (A circuit with N junctions has N-1 independent junction equations).
4. Apply Kirchhoff's Loop Rule to each independent closed loop in the circuit.
5. Solve the system of linear equations obtained from the Junction and Loop Rules to find the unknown currents.

Wheatstone Bridge (Condition for balance)

The Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. The primary advantage of using a bridge circuit is that it allows for very precise measurements using a null indicator (like a galvanometer), rather than relying on the absolute accuracy of voltmeters or ammeters.

The basic Wheatstone bridge circuit consists of four resistive arms ($P, Q, R, S$) connected to form a quadrilateral ABCD. A source of EMF (like a battery) is connected across two opposite vertices (say, A and C). A galvanometer (G) is connected across the other two opposite vertices (say, B and D).

Wheatstone Bridge circuit.

Typically, three of the resistances (say, P, Q, R) are known, and the fourth resistance (S) is unknown. One of the known resistances (usually R) is made variable (e.g., using a resistance box).

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Working Principle and Balance Condition

The working principle involves adjusting the variable resistance R until the galvanometer shows zero deflection. When the galvanometer shows zero deflection, it means there is no current flowing through the galvanometer wire BD. This condition is called the balance condition of the Wheatstone bridge.

When the bridge is balanced, the potential at point B must be equal to the potential at point D ($V_B = V_D$). This means there is no potential difference across the galvanometer.

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Derivation of the Balance Condition ($ P/Q = R/S $)

Let the current from the battery be $I$. At junction A, this current divides into $I_1$ flowing through P and $I_2$ flowing through R. At junction B, if the galvanometer shows zero deflection, the current through the galvanometer ($I_g$) is zero. This means all the current $I_1$ that flows through P must flow through Q, and all the current $I_2$ that flows through R must flow through S.

So, when balanced:

• Current through P = Current through Q = $I_1$
• Current through R = Current through S = $I_2$

Since $V_B = V_D$, the potential difference across arm AB ($V_A - V_B$) must be equal to the potential difference across arm AD ($V_A - V_D$).

$ V_A - V_B = V_A - V_D \implies V_{AB} = V_{AD} $

Using Ohm's Law:

$ I_1 P = I_2 R \quad \ldots (1) $

Similarly, the potential difference across arm BC ($V_B - V_C$) must be equal to the potential difference across arm DC ($V_D - V_C$).

$ V_B - V_C = V_D - V_C \implies V_{BC} = V_{DC} $

Using Ohm's Law:

$ I_1 Q = I_2 S \quad \ldots (2) $

Now, divide equation (1) by equation (2):

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

Assuming $I_1 \neq 0$ and $I_2 \neq 0$ (which is true unless the total current is zero), we can cancel $I_1$ and $I_2$:

$ \frac{P}{Q} = \frac{R}{S} $

or equivalently, $ P S = Q R $

This is the balance condition for a Wheatstone bridge. If this ratio of resistances is maintained, no current will flow through the galvanometer arm.

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Using the Wheatstone Bridge to Measure Unknown Resistance

To measure an unknown resistance $S$, one typically sets up the bridge with three known resistors $P, Q, R$. $R$ is made variable using a resistance box. The unknown resistance $S$ is connected in the fourth arm. The variable resistance $R$ is adjusted until the galvanometer shows zero deflection. At this point, the bridge is balanced. The value of the variable resistance $R$ is noted, along with the fixed resistances $P$ and $Q$.

Using the balance condition $P/Q = R/S$, the unknown resistance $S$ can be calculated:

$ S = R \frac{Q}{P} $

The ratio $Q/P$ is often fixed (e.g., $1:1$, $10:1$, $1:10$), and $R$ is adjusted to find the null point.

The Wheatstone bridge provides a very accurate method for measuring resistance because the measurement relies on finding a null point (zero current), which can be detected with high sensitivity using a galvanometer. The accuracy is primarily limited by the sensitivity of the galvanometer and the precision of the known resistances and the variable resistance.

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Answer:

Meter Bridge

The Meter Bridge is a practical application of the Wheatstone bridge circuit, designed specifically for measuring an unknown electrical resistance. It uses a 1-meter long wire of uniform material and cross-sectional area as two of the resistive arms of the bridge.

Meter Bridge setup. R is known resistance, X is unknown resistance, G is galvanometer, J is jockey.

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Construction

A typical Meter Bridge apparatus consists of:

• A wooden board with two L-shaped thick copper strips and one straight copper strip fixed on it. Three gaps are formed: a left gap, a right gap, and a central gap.
• A uniform wire of 1-meter length, usually made of materials like manganin or constantan (alloys chosen for their constant resistivity with temperature and relatively high resistance). The wire is stretched tightly along a metre scale fixed on the board, connecting the ends of the two L-shaped copper strips.
• Binding posts on the copper strips for connecting external resistances and the galvanometer.
• A galvanometer (G) and a jockey (J). The galvanometer is connected to the central copper strip, and the other terminal of the galvanometer is connected to the jockey. The jockey is a sliding contact that can be moved along the meter wire.

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Working Principle and Measurement of Unknown Resistance

The Meter Bridge works on the principle of the balanced Wheatstone bridge. The unknown resistance ($X$) is connected in one gap (say, the right gap), and a known resistance ($R$) from a resistance box is connected in the other gap (the left gap). A battery (source of EMF) is connected across the ends of the meter wire (points A and C). The galvanometer is connected between the central copper strip (point B) and the jockey (point D) that can touch the wire at any point.

The jockey is moved along the wire (AC) until a point D is found where the galvanometer shows zero deflection. This is the null point or balance point. At the balance point, there is no current through the galvanometer, indicating that the potential at point B is equal to the potential at point D ($V_B = V_D$). The bridge is then balanced.

In this setup, the four arms of the Wheatstone bridge are:

• Arm AB: Resistance $R$ (from the resistance box) - This corresponds to resistance P in the standard Wheatstone bridge diagram (or R).
• Arm BC: Unknown resistance $X$ - This corresponds to resistance Q (or S).
• Arm AD: The resistance of the wire segment of length $l_1$ - This corresponds to resistance R (or P).
• Arm DC: The resistance of the wire segment of length $l_2 = (100 - l_1)$ cm - This corresponds to resistance S (or Q).

Let $\rho_w$ be the resistance per unit length of the meter wire. The resistance of segment AD is $R_{AD} = \rho_w l_1$, and the resistance of segment DC is $R_{DC} = \rho_w l_2 = \rho_w (100 - l_1)$.

Applying the balance condition of the Wheatstone bridge: $\frac{\text{Resistance in left gap}}{\text{Resistance in right gap}} = \frac{\text{Resistance of wire segment under left gap}}{\text{Resistance of wire segment under right gap}}$

$ \frac{R}{X} = \frac{R_{AD}}{R_{DC}} = \frac{\rho_w l_1}{\rho_w (100 - l_1)} $

The resistance per unit length $\rho_w$ cancels out:

$ \frac{R}{X} = \frac{l_1}{100 - l_1} $

Rearranging to find the unknown resistance $X$:

$ X = R \frac{100 - l_1}{l_1} $

By measuring the balancing length $l_1$ (the distance of the null point from end A) and knowing the standard resistance $R$, the unknown resistance $X$ can be accurately determined. The length $l_1$ is read directly from the meter scale.

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Importance of Uniform Wire and Jockey Use

The accuracy of the meter bridge depends on the wire being uniform in material and cross-sectional area throughout its length. Non-uniformity would mean that the resistance is not linearly proportional to the length, invalidating the formula.

The jockey should only be touched gently to the wire at the point where the balance is sought. Sliding the jockey can scrape the wire, altering its cross-sectional area and thus its uniformity.

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Answer:

Potentiometer (Comparison of EMF, Measurement of Internal Resistance)

A potentiometer is a versatile electrical instrument used primarily for accurate measurement of potential differences. It is also used to compare the EMFs of cells and to determine the internal resistance of a cell. Unlike a voltmeter, which draws some current from the circuit being measured, a potentiometer draws no current from the source at the point of measurement (at balance), making it an ideal device for measuring EMF.

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Principle of Potentiometer

The working principle of a potentiometer is based on the fact that the potential drop across any portion of a wire of uniform cross-sectional area and uniform composition, carrying a constant current, is directly proportional to the length of that portion of the wire.

Basic Potentiometer setup: Primary circuit establishing potential gradient.

Consider a uniform wire AB of length $L$ and resistance $R_w$. A battery with EMF $\mathcal{E}_p$ (driver cell) and internal resistance $r_p$, along with a rheostat (variable resistance) $R_{ext}$ and a key, is connected across the wire AB. This forms the primary circuit.

The current ($I$) flowing through the wire AB is given by:

$ I = \frac{\mathcal{E}_p}{R_w + R_{ext} + r_p} $

Assuming $R_{ext}$ is adjusted to keep the current constant, and the wire is uniform ($\rho_w$ is resistance per unit length), the potential drop across the entire length $L$ of the wire is $V_{AB} = I R_w = I (\rho_w L)$.

The potential drop per unit length of the wire is called the potential gradient ($k$):

$ k = \frac{V_{AB}}{L} = \frac{IR_w}{L} = I \rho_w $

Since $I$ and $\rho_w$ are constant, the potential gradient $k$ is constant along the wire.

The potential difference across any length $l$ of the wire from end A is:

$ V_l = k l $

Thus, $V_l \propto l$, which is the principle of the potentiometer.

In the secondary circuit, the cell or component whose potential difference is to be measured is connected in series with a galvanometer and a jockey. One terminal of the cell is connected to end A of the potentiometer wire (the high potential end of the primary circuit), and the other terminal (through the galvanometer and jockey) is connected to a point D on the wire. The jockey is moved along the wire to find a point D where the galvanometer shows zero deflection. This is the null point or balance point.

At the null point, no current flows through the secondary circuit (the cell and the galvanometer). This means the EMF of the cell (or the potential difference being measured) is exactly equal to the potential drop across the length of the potentiometer wire from A to D ($l$).

$ \text{Potential difference measured} = k l $

The great advantage is that at balance, the cell or component being measured is not supplying any current, so the measured voltage is its true EMF (for a cell) or the true potential difference without any drop due to internal resistance or external load.

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Comparison of EMFs of Two Cells

The potentiometer can be used to compare the EMFs of two different cells, say with EMFs $\mathcal{E}_1$ and $\mathcal{E}_2$.

Potentiometer setup for comparing EMFs. $K_1$ is a two-way key.

Procedure:

1. Set up the primary circuit with a stable driver cell and a potentiometer wire. Ensure a constant current flows.
2. Connect the positive terminals of both cells ($\mathcal{E}_1$ and $\mathcal{E}_2$) to the end A of the potentiometer wire (the positive terminal of the driver cell connection).
3. Use a two-way key ($K_1$) to connect the negative terminal of $\mathcal{E}_1$ to the galvanometer, and the other end of the galvanometer to the jockey.
4. Find the null point D for cell $\mathcal{E}_1$ by moving the jockey. Let the balancing length from A be $l_1$. At balance, the EMF of the cell is equal to the potential drop across length $l_1$:

$ \mathcal{E}_1 = k l_1 \quad \ldots (1) $

5. Now, without changing the current in the primary circuit, connect the negative terminal of $\mathcal{E}_2$ to the galvanometer using the two-way key.
6. Find the new null point D' for cell $\mathcal{E}_2$ by moving the jockey. Let the balancing length from A be $l_2$. At balance:

$ \mathcal{E}_2 = k l_2 \quad \ldots (2) $

Derivation: Divide equation (1) by equation (2):

$ \frac{\mathcal{E}_1}{\mathcal{E}_2} = \frac{k l_1}{k l_2} $

$ \frac{\mathcal{E}_1}{\mathcal{E}_2} = \frac{l_1}{l_2} $

Thus, the ratio of the EMFs of the two cells is equal to the ratio of their corresponding balancing lengths. By measuring $l_1$ and $l_2$, the ratio $\mathcal{E}_1/\mathcal{E}_2$ can be determined. If one of the EMFs (say $\mathcal{E}_2$) is known (a standard cell), the other EMF ($\mathcal{E}_1$) can be calculated.

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Measurement of Internal Resistance of a Cell ($ r $)

The potentiometer can also be used to determine the internal resistance of a given cell (with EMF $\mathcal{E}$ and internal resistance $r$).

Potentiometer setup for measuring internal resistance. R is a resistance box, $K_1, K_2$ are keys.

Procedure:

1. Set up the primary circuit with a stable current.
2. Connect the cell whose internal resistance is to be measured in the secondary circuit. Connect its positive terminal to end A of the potentiometer wire. Connect the negative terminal through a key $K_1$ and a galvanometer to the jockey.
3. Connect a resistance box ($R$) in parallel with the cell, with a key $K_2$ in series with the resistance box.

4. Step 1: Key $K_2$ is Open. When key $K_2$ is open, no current flows through the external resistance $R$. The current from the cell itself is essentially zero when balanced by the potentiometer. The potential difference balanced by the potentiometer is the EMF ($\mathcal{E}$) of the cell.

   Find the null point D$_1$ by moving the jockey (with $K_1$ closed). Let the balancing length be $l_1$. At balance:

   $ \mathcal{E} = k l_1 \quad \ldots (3) $

5. Step 2: Key $K_2$ is Closed. Close key $K_2$. Now, current ($I$) flows from the cell through the external resistance $R$ taken from the resistance box. The potential difference across the external resistance $R$ (which is also the terminal voltage $V$ of the cell) is balanced by the potentiometer.

   Find the new null point D$_2$ by moving the jockey (with $K_1$ still closed). Let the balancing length be $l_2$. At balance, the terminal voltage $V$ is equal to the potential drop across length $l_2$:

   $ V = k l_2 \quad \ldots (4) $

Derivation: We know the relationship between EMF ($\mathcal{E}$), terminal voltage ($V$), current ($I$), and internal resistance ($r$):

$ V = \mathcal{E} - Ir $

Also, the current flowing through the external resistance $R$ is $I = V/R$.

Substitute $I = V/R$ into the equation for $V$:

$ V = \mathcal{E} - \left(\frac{V}{R}\right) r $

Now, substitute the expressions for $\mathcal{E}$ and $V$ from the potentiometer balancing lengths (equations 3 and 4):

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

Assuming $k \neq 0$, we can divide the entire equation by $k$:

$ l_2 = l_1 - \frac{l_2}{R} r $

Rearrange to solve for $r$:

$ \frac{l_2}{R} r = l_1 - l_2 $

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

$ r = R \left( \frac{l_1}{l_2} - 1 \right) $

By measuring the balancing lengths $l_1$ (for open circuit, $\mathcal{E}$) and $l_2$ (for closed circuit with external resistance $R$), the internal resistance $r$ of the cell can be calculated using this formula.

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