Q.Let there be $n$ resistors $R_1 \ldots R_n$ with $R_{max} = \max(R_1 \ldots R_n)$ and $R_{min} = \min\{R_1 \ldots R_n\}$. Show that when they are connected in parallel, the resultant resistance $R_P < R_{min}$ and when they are connected in series, the resultant resistance $R_S > R_{max}$. Interpret the result physically.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.When two resistors are connected in the two gaps of a meter bridge, the balancing point is obtained at 25 cm from the left end of the bridge wire. When a $48~\Omega$ resistor is connected in series to the smaller of the two resistors, the balancing point is obtained at 75 cm from the left end of the bridge wire. Then the initial values of the resistors connected in the left and right gaps of the bridge are respectively (A) $18~\Omega,~54~\Omega$ (B) $6~\Omega,~18~\Omega$ (C) $12~\Omega,~36~\Omega$ (D) $24~\Omega,~72~\Omega$

   ›Reveal solutionSolution

   This problem involves applying the meter bridge principle twice. We use the balancing point information to set up two equations relating the unknown resistances and then solve them simultaneously. The initial values of the resistors are $6~\Omega$ and $18~\Omega$.

   Concept and Intuition

   The meter bridge is a practical application of the Wheatstone bridge, designed to measure an unknown electrical resistance. It consists of a uniform wire of 1-meter length (100 cm) stretched along a scale, and two gaps where resistors can be connected. A galvanometer is connected between a central terminal and a sliding jockey that can touch any point on the bridge wire.

   The fundamental principle is to find a "balancing point" on the wire where the galvanometer shows zero deflection. When the galvanometer reads zero, it means no current flows through it, indicating that the potential difference across its terminals is zero. This condition signifies that the Wheatstone bridge is balanced.

   At the balancing point, the ratio of the resistances in the two gaps is equal to the ratio of the lengths of the wire segments on either side of the balancing point. This is because the resistance of a uniform wire is directly proportional to its length.

   For a balanced meter bridge, if $R_L$ is the resistance in the left gap, $R_R$ is the resistance in the right gap, and $l$ is the balancing length from the left end of the bridge wire, then:

   $\frac{R_L}{R_R} = \frac{l}{100 - l}$

   Step-by-step solution

   1. Analyze the first scenario: Let the initial resistance in the left gap be $R_L$ and in the right gap be $R_R$. The problem states that the balancing point is obtained at $l_1 = 25$ cm from the left end. Using the meter bridge formula:

   $\frac{R_L}{R_R} = \frac{l_1}{100 - l_1}$

   Substitute the given balancing length:
   

   $\frac{R_L}{R_R} = \frac{25}{100 - 25}$

   $\frac{R_L}{R_R} = \frac{25}{75}$

   $\frac{R_L}{R_R} = \frac{1}{3}$

   This gives us our first relationship between $R_L$ and $R_R$:
   

   $R_R = 3R_L \quad \text{(Equation 1)}$

   2. Identify the smaller resistor:

      From Equation 1, since $R_R = 3R_L$, it is clear that $R_L$ is the smaller of the two initial resistors. This identification is crucial because the next step involves connecting an additional resistor to the smaller one.

   3. Analyze the second scenario:

      A $48~\Omega$ resistor is connected in series to the smaller resistor, which we identified as $R_L$.

      So, the new resistance in the left gap, $R_L'$, becomes $R_L + 48~\Omega$.

      The resistance in the right gap remains $R_R$.

      The new balancing point is obtained at $l_2 = 75$ cm from the left end.

      Applying the meter bridge formula again with the new values:

   $\frac{R_L'}{R_R} = \frac{l_2}{100 - l_2}$

   Substitute $R_L' = R_L + 48$ and $l_2 = 75$:
   $$ \frac{R_L + 48}{R_R} = \frac{75}{100 - 75} $$ …
   

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.In a meter bridge with resistors $R_1$ and $R_2$ in the left and right gaps respectively, the balancing point is obtained at 25 cm from the left end of the bridge wire. If shunt resistances of $2~\Omega$ each are connected to both $R_1$ and $R_2$, the balancing point shifts by 15 cm. The initial values of $R_1$ and $R_2$ respectively are (A) $2~\Omega,~6~\Omega$ (B) $1~\Omega,~3~\Omega$ (C) $4~\Omega,~12~\Omega$ (D) $5~\Omega,~15~\Omega$

   ›Reveal solutionSolution

   The meter bridge balances when the ratio of resistances equals the ratio of wire lengths. Adding identical shunts changes both resistances in parallel, shifting the balance point; solving the two resulting equations gives $R_1 = 1\,\Omega$ and $R_2 = 3\,\Omega$, so option (B) is correct.

   Concept & Intuition

   A meter bridge is a Wheatstone bridge with a uniform wire. At balance, the ratio of the two unknown resistances equals the ratio of the lengths from the left end to the balance point and from the balance point to the right end. When we add a shunt (a resistor in parallel) to each arm, each resistance decreases. The new balance point shifts because the ratio of the two modified resistances is different. The problem gives us the original balance length (25 cm) and the shift (15 cm) — we must decide whether the shift is toward the left or right. Since adding a shunt reduces each resistance, the ratio changes; the direction of shift tells us which arm’s resistance decreased more proportionally. We’ll set up equations for both cases and see which yields positive, sensible resistances.

   Step-by-step solution

   1. Original balance condition In a meter bridge, if the balance point is at $L$ cm from the left end, then

   $\frac{R_1}{R_2} = \frac{L}{100 - L}$

   Here $L = 25$ cm, so

   $\frac{R_1}{R_2} = \frac{25}{75} = \frac{1}{3} \quad \Rightarrow \quad R_2 = 3R_1.$

   2. Effect of shunts A shunt of $2\,\Omega$ connected in parallel with $R_1$ gives an effective resistance

   $R_1' = \frac{R_1 \times 2}{R_1 + 2}.$

   Similarly,

   $R_2' = \frac{R_2 \times 2}{R_2 + 2}.$

   3. New balance condition

      The balance point shifts by 15 cm. It could shift toward the left (new length $25 - 15 = 10$ cm) or toward the right (new length $25 + 15 = 40$ cm). We must test both.

      • Case A (shift to left, $L' = 10$ cm):

   $\frac{R_1'}{R_2'} = \frac{10}{90} = \frac{1}{9}.$

    Substitute $R_2 = 3R_1$ into $R_2'$:  
   

   $R_2' = \frac{3R_1 \times 2}{3R_1 + 2} = \frac{6R_1}{3R_1 + 2}.$

    Then  
   

   $\frac{R_1'}{R_2'} = \frac{\frac{2R_1}{R_1+2}}{\frac{6R_1}{3R_1+2}} = \frac{2R_1}{R_1+2} \cdot \frac{3R_1+2}{6R_1} = \frac{3R_1+2}{3(R_1+2)}.$

    Set equal to $1/9$:  
   

   $\frac{3R_1+2}{3(R_1+2)} = \frac{1}{9} \quad \Rightarrow \quad 9(3R_1+2) = 3(R_1+2).$

    Simplify: $27R_1 + 18 = 3R_1 + 6 \Rightarrow 24R_1 = -12 \Rightarrow R_1 = -0.5\,\Omega$.  
    Negative resistance is impossible, so this case is invalid.
   

   • Case B (shift to right, $L' = 40$ cm): $\frac{R_1'}{R_2'} = \frac{40}{60} = \frac{2}{3}.$ …

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3. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.In a potentiometer experiment, a wire of length 10 m and resistance $5~\Omega$ is connected to a cell of emf $2.2~\text{V}$. If the potential difference between two points separated by a distance of $660~\text{cm}$ on potentiometer wire is $1.1~\text{V}$, then the internal resistance of the cell is (A) $1.6~\Omega$ (B) $1.4~\Omega$ (C) $1.2~\Omega$ (D) $1~\Omega$

   ›Reveal solutionSolution

   Internal resistance $r = 1.6\,\Omega$ — option (A).

   Given: wire length $L = 10\,\text{m}$, wire resistance $R = 5\,\Omega$, driver cell emf $E = 2.2\,\text{V}$; the fall across $\ell = 660\,\text{cm} = 6.6\,\text{m}$ is $V = 1.1\,\text{V}$.

   Step 1 — Current in the potentiometer wire.

   The cell (emf $E$, internal resistance $r$) drives the wire:

   $I = \frac{E}{R + r} = \frac{2.2}{5 + r}.$

   Step 2 — Potential drop over the 6.6 m length. …

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4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The area of cross-section of a potentiometer wire is $6 \times 10^{-7} \ \mathrm{m}^2$. The potential difference per unit length of the potentiometer wire when it is connected to a cell of negligible internal resistance and a resistor in series is $0.15 \ \mathrm{Vm}^{-1}$. If the current through potentiometer wire is $0.3 \ \mathrm{A}$, then the resistivity of the material of the potentiometer wire is (A) $4 \times 10^{-6} \ \Omega \mathrm{m}$ (B) $4 \times 10^{-7} \ \Omega \mathrm{m}$ (C) $3 \times 10^{-6} \ \Omega \mathrm{m}$ (D) $3 \times 10^{-7} \ \Omega \mathrm{m}$

   ›Reveal solutionSolution

   The resistivity is found by relating the potential gradient to the current and cross-sectional area via $R = \rho L / A$ and $V/L = IR/L$. The result is $\rho = 3 \times 10^{-7} \ \Omega \mathrm{m}$, option (D).

   The key idea here is that the potential difference per unit length (the potential gradient) along the potentiometer wire is directly linked to the wire’s resistance per unit length. Since the wire carries a steady current, Ohm’s law in its local form — $V = IR$ — applies, and the resistance of a length $L$ of wire is $R = \rho L / A$. Combining these gives a clean expression for resistivity.

   Let’s work through it step by step.

   1. Write down what’s given.

      Cross-sectional area: $A = 6 \times 10^{-7} \ \mathrm{m}^2$

      Potential gradient: $\frac{V}{L} = 0.15 \ \mathrm{Vm}^{-1}$

      Current: $I = 0.3 \ \mathrm{A}$

      The cell has negligible internal resistance, so the current is steady and determined only by the external resistor and the wire itself.

   2. Relate potential gradient to resistance per unit length.

      For a uniform wire, the potential difference across a length $L$ is $V = IR$, where $R$ is the resistance of that length. So

   $\frac{V}{L} = I \cdot \frac{R}{L}.$

   Here $\frac{R}{L}$ is the resistance per unit length of the wire.

   3. Express resistance per unit length in terms of resistivity. The resistance of a length $L$ is $R = \rho \frac{L}{A}$, so

   $\frac{R}{L} = \frac{\rho}{A}.$

   This is a standard result: for a given material and cross-section, resistance per unit length is constant.

   4. Combine the equations. Substitute $\frac{R}{L} = \rho / A$ into the potential gradient equation:

   $\frac{V}{L} = I \cdot \frac{\rho}{A}.$

   Rearranging for $\rho$:

   $\rho = \frac{A}{I} \cdot \frac{V}{L}.$

   5. Plug in the numbers. $$\rho = \frac{6 \times 10^{-7}}{0.3} \times 0.15. …$$

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5. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Three resistors $R_1$, $R_2$ and $R_3$ when connected in parallel to a battery of negligible internal resistance, the currents through the three resistors are $2\ \mathrm{A}$, $3\ \mathrm{A}$ and $6\ \mathrm{A}$ respectively. If the resistors are connected in series to the same battery, the current in the circuit is (A) $1\ \mathrm{A}$ (B) $2\ \mathrm{A}$ (C) $3\ \mathrm{A}$ (D) $11\ \mathrm{A}$

   ›Reveal solutionSolution

   In parallel, voltage is the same across all resistors, so the currents tell us the conductances. In series, the same battery drives current through the total series resistance. The answer is 1 A.

   The key idea is that the battery voltage is fixed and the internal resistance is negligible, so the battery behaves as an ideal voltage source. In a parallel connection, each resistor sees the same voltage $V$. The current through a resistor is $I = V/R$, so the current is inversely proportional to the resistance. That means the currents $2\ \mathrm{A}$, $3\ \mathrm{A}$, and $6\ \mathrm{A}$ directly give us the relative conductances (reciprocals of resistance). In series, the same battery voltage drives current through the sum of the resistances. We can find the total series resistance from the parallel data without ever needing the actual value of $V$.

   1. Find the resistances from the parallel currents. Let the battery voltage be $V$. Then

   $R_1 = \frac{V}{2},\quad R_2 = \frac{V}{3},\quad R_3 = \frac{V}{6}.$

   Notice that the smallest current corresponds to the largest resistance, which makes sense.

   2. Compute the total series resistance. When connected in series, the total resistance is

   $R_s = R_1 + R_2 + R_3 = \frac{V}{2} + \frac{V}{3} + \frac{V}{6}.$

   Find a common denominator (6):

   $R_s = \frac{3V}{6} + \frac{2V}{6} + \frac{V}{6} = \frac{6V}{6} = V.$

   So the series combination has a resistance exactly equal to $V$ ohms (if $V$ is in volts).

   3. Find the series current. The same battery of voltage $V$ is now applied across $R_s = V$. Using Ohm’s law:

   $I_{\text{series}} = \frac{V}{R_s} = \frac{V}{V} = 1\ \mathrm{A}.$

   The voltage cancels out completely — the answer does not depend on the actual value of $V$. …

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6. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.When two resistors of resistances $(123 \pm 2)\ \Omega$ and $(227 \pm 4)\ \Omega$ are connected in series, then the value of equivalent resistance is (A) $(350 \pm 7)\ \Omega$ (B) $(350 \pm 1)\ \Omega$ (C) $(350 \pm 12)\ \Omega$ (D) $(350 \pm 3)\ \Omega$

   ›Reveal solutionSolution

   When resistors are connected in series, their nominal resistances add up, and their absolute uncertainties also add up. For the given resistors, the equivalent resistance is $(350 \pm 6)\ \Omega$. However, in some contexts, uncertainties are rounded up to the next integer to ensure a conservative estimate, leading to $(350 \pm 7)\ \Omega$.

   In physics, when we measure quantities, there's always some degree of uncertainty. When we combine these quantities, their uncertainties also combine. This problem asks us to find the equivalent resistance of two resistors connected in series, along with the uncertainty in that equivalent resistance.

   The core concepts here are:

   1. Resistors in Series: When resistors are connected in series, the total (equivalent) resistance is simply the sum of their individual resistances. This is because the current flows through each resistor sequentially, encountering each resistance in turn.
   2. Propagation of Uncertainties (for Addition/Subtraction): When two quantities, say $A \pm \Delta A$ and $B \pm \Delta B$, are added or subtracted to get a resultant quantity $X = A \pm B$, the maximum possible absolute uncertainty in $X$ is the sum of the individual absolute uncertainties. That is, $\Delta X = \Delta A + \Delta B$. This rule ensures that the calculated range for $X$ covers all possible values given the uncertainties in $A$ and $B$.

   Let's apply these concepts to the given problem.

   1. Identify the given values:

      We are given two resistors with their resistances and associated absolute uncertainties:

      • Resistance of the first resistor, $R_1 = (123 \pm 2)\ \Omega$. Here, the nominal resistance is $R_{1, \text{nom}} = 123\ \Omega$, and its absolute uncertainty is $\Delta R_1 = 2\ \Omega$.
      • Resistance of the second resistor, $R_2 = (227 \pm 4)\ \Omega$. Here, the nominal resistance is $R_{2, \text{nom}} = 227\ \Omega$, and its absolute uncertainty is $\Delta R_2 = 4\ \Omega$.

   2. Calculate the nominal equivalent resistance:

      For resistors connected in series, the equivalent resistance $R_{eq}$ is the sum of the individual resistances.

   $R_{eq, \text{nom}} = R_{1, \text{nom}} + R_{2, \text{nom}}$

   Substituting the nominal values:
   

   $R_{eq, \text{nom}} = 123\ \Omega + 227\ \Omega = 350\ \Omega$

   3. Calculate the absolute uncertainty in the equivalent resistance: When quantities are added, their absolute uncertainties add up.

   $\Delta R_{eq} = \Delta R_1 + \Delta R_2$

   Substituting the absolute uncertainties:
   

   $\Delta R_{eq} = 2\ \Omega + 4\ \Omega = 6\ \Omega$

   4. Express the final result and consider options: Based on our calculations, the equivalent resistance is $(350 \pm 6)\ \Omega$. Now, let's look at the given options: (A) $(350 \pm 7)\ \Omega$ (B) $(350 \pm 1)\ \Omega$ (C) $(350 \pm 12)\ \Omega$ …

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7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The error in the measurement of resistance, when $(10 \pm 0.5)\ \mathrm{A}$ current passing through it produces a potential difference of $(100 \pm 6)\ \mathrm{V}$ across it is (A) $1\ \%$ (B) $5.5\ \%$ (C) $6.5\ \%$ (D) $11\ \%$

   ›Reveal solutionSolution

   Resistance $R=V/I$, so the percentage errors add: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = 6\% + 5\% = 11\%$ — option (D).

   Step-by-step solution

   By Ohm's law, $R = \dfrac{V}{I}$. For a quotient, fractional errors add:

   $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}.$

   Substitute the measured values $V = (100 \pm 6)\ \text{V}$ and $I = (10 \pm 0.5)\ \text{A}$: …

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8. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Three identical rods are joined as shown in the figure. The left and right ends are kept at $0^\circ \mathrm{C}$ and $90^\circ \mathrm{C}$ as shown in the figure. The temperature $\theta$ at the junction of the rods is (A) $60^\circ \mathrm{C}$ (B) $45^\circ \mathrm{C}$ (C) $30^\circ \mathrm{C}$ (D) $20^\circ \mathrm{C}$

   ›Reveal solutionSolution

   In the steady state the heat arriving at the junction from the two $90^\circ\mathrm{C}$ rods must equal the heat leaving through the single $0^\circ\mathrm{C}$ rod. Since the rods are identical this gives $2(90-\theta)=\theta$, so $\theta = 60^\circ\mathrm{C}$ — option (A).

   The concept first

   Heat conduction along a rod is governed by

   $\frac{dQ}{dt} = \frac{kA}{L}\,\Delta T$

   where $k$ is thermal conductivity, $A$ the cross-section, $L$ the length and $\Delta T$ the temperature difference between the ends. The combination $\dfrac{kA}{L}$ is the thermal conductance — the exact analogue of electrical conductance $1/R$, with temperature playing the role of potential and heat current the role of electric current.

   That analogy is the key. Steady state means the junction's temperature is no longer changing, so it is neither storing nor losing energy — which means

   $\sum \text{heat in} = \sum \text{heat out}$

   exactly like Kirchhoff's junction rule for currents. Note the danger of a lazy guess: many students average $0$ and $90$ and say $45^\circ\mathrm{C}$. That would be right if one rod were at $90^\circ\mathrm{C}$ and one at $0^\circ\mathrm{C}$. Here two rods pump heat in from the hot side, so the junction must sit hotter than the midpoint.

   Step-by-step

   1. Define the common conductance. All three rods are identical, so

   $C = \frac{kA}{L} \quad \text{is the same for each.}$

   2. Heat currents into the junction (from the two rods whose far ends are at $90^\circ\mathrm{C}$). Heat flows from hot to cold, and the junction (at $\theta$) is cooler than $90^\circ\mathrm{C}$: $H_{\text{in}} = C(90-\theta) + C(90-\theta) = 2C(90-\theta).$ …

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9. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The relation between the current i (in ampere) in a conductor and the time t (in second) is $i = 12t + 9t^2$. The charge passing through the conductor between the times $t = 2\,s$ and $t = 10\,s$ is (A) 3720 C (B) 3648 C (C) 3600 C (D) 3552 C

   ›Reveal solutionSolution

   Charge is $\int i\,dt$ from $2\,\text{s}$ to $10\,\text{s}$, giving $3552\,\text{C}$.

   Current: $i = 12t + 9t^{2}$.

   $q = \int_{2}^{10} (12t + 9t^{2})\,dt = \left[\,6t^{2} + 3t^{3}\,\right]_{2}^{10}.$

   At $t=10$: $6(100) + 3(1000) = 600 + 3000 = 3600$. …

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10. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
    Q.In a potentiometer experiment, a cell of emf $1.5\ \mathrm{V}$ connected in the secondary circuit gives a balancing length of $165\ \mathrm{cm}$. If a resistance of $5\ \Omega$ is connected parallel to the cell, the balancing length of the wire is $150\ \mathrm{cm}$. The internal resistance of the cell is (A) $5\ \Omega$ (B) $1.5\ \Omega$ (C) $1\ \Omega$ (D) $0.5\ \Omega$

    ›Reveal solutionSolution

    The balancing length is proportional to the terminal voltage across the cell. By comparing the open-circuit and loaded cases, the internal resistance is found to be $0.5\ \Omega$.

    The key idea here is that a potentiometer measures the terminal voltage of a cell, not its emf directly — unless the cell is on open circuit. When you connect a resistor across the cell, current flows through its internal resistance, causing a voltage drop inside the cell. The balancing length is proportional to the voltage actually available at the terminals.

    Let’s work through it.

    1. Open-circuit case (no external resistor) The cell’s emf $E = 1.5\ \mathrm{V}$ gives a balancing length $L_1 = 165\ \mathrm{cm}$. Since no current flows through the cell, the terminal voltage equals the emf. The potential gradient $k$ (volts per cm) along the wire is constant, so

    $E = k L_1 \quad \Rightarrow \quad 1.5 = k \times 165.$

    2. Loaded case (resistor $R = 5\ \Omega$ in parallel) When a $5\ \Omega$ resistor is connected across the cell, the cell delivers current $I$. The terminal voltage $V$ is now less than $E$ because of the drop $Ir$ across the internal resistance $r$:

    $V = E - Ir.$

    This $V$ gives a new balancing length $L_2 = 150\ \mathrm{cm}$, so

    $V = k L_2.$

    3. Relating the two cases

       From step 1, $k = \frac{1.5}{165}$.

       From step 2, $V = \frac{1.5}{165} \times 150 = 1.5 \times \frac{150}{165} = 1.5 \times \frac{10}{11} = \frac{15}{11}\ \mathrm{V}.$

    4. Finding the current $I$

       The external resistance is $5\ \Omega$ (connected directly across the cell).

       The terminal voltage $V$ is also the voltage across this $5\ \Omega$ resistor, so

       $I = \frac{V}{R} = \frac{15/11}{5} = \frac{15}{55} = \frac{3}{11}\ \mathrm{A}.$ …

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11. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.Two parallel long straight wires carry currents $10\,\mathrm{A}$ and $I$ ($>10\,\mathrm{A}$). If the currents in the wires are in the same direction, the magnetic field at a point equidistant between the conductors is $10^{-5}\,\mathrm{T}$. If the currents in the wires are in the opposite directions, the magnetic field at a point equidistant between the conductors is $3 \times 10^{-5}\,\mathrm{T}$. The value of $I$ is (A) $20\,\mathrm{A}$ (B) $15\,\mathrm{A}$ (C) $5\,\mathrm{A}$ (D) $12\,\mathrm{A}$

    ›Reveal solutionSolution

    The net magnetic field between two parallel wires depends on whether currents are parallel (fields oppose) or antiparallel (fields add). Using the formula $B = \frac{\mu_0 I}{2\pi d}$ and the given fields, we solve two equations to find $I = 20\,\mathrm{A}$.

    The key concept here is the magnetic field due to a long straight current-carrying wire. At a perpendicular distance $r$ from the wire, the field magnitude is $B = \frac{\mu_0 I}{2\pi r}$, and its direction is given by the right-hand rule. When two parallel wires carry current, the fields at a point between them either oppose or reinforce depending on the current directions.

    Let the distance between the wires be $2d$, so the point exactly midway is at distance $d$ from each wire. The field due to a single wire at that point is $B_0 = \frac{\mu_0 I}{2\pi d}$.

    1. Case 1: Currents in the same direction At the midpoint, the fields from the two wires point in opposite directions (right-hand rule). The net field is the difference:

    $B_{\text{same}} = \frac{\mu_0}{2\pi d}(I - 10) = 10^{-5}\,\mathrm{T}$

    (since $I > 10\,\mathrm{A}$, the field from the larger current dominates).

    2. Case 2: Currents in opposite directions Now the fields at the midpoint point in the same direction, so they add:

    $B_{\text{opp}} = \frac{\mu_0}{2\pi d}(I + 10) = 3 \times 10^{-5}\,\mathrm{T}$

    3. Solve the system Divide the second equation by the first: …

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12. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.In the given network, if the charges on the capacitors in steady state are same, then $\left(\dfrac{C_1}{C_2}\right)$ is (A) $5:3$ (B) $3:5$ (C) $1:1$ (D) $1:3$

    ›Reveal solutionSolution

    In steady state, no current flows through capacitors. Equal charges on $C_1$ and $C_2$ means $C_1 V_1 = C_2 V_2$. The voltage division across the resistors in series determines the capacitor voltages, giving $\dfrac{C_1}{C_2} = \boxed{3:5}$.

    When a DC circuit reaches steady state, capacitors act as open circuits because no current flows through them. This means we can analyze the resistor network alone to find the voltages across each capacitor.

    The key insight: if two capacitors carry the same charge $Q$, then $Q = C_1 V_1 = C_2 V_2$, which immediately tells us that the capacitance ratio is inversely proportional to the voltage ratio:

    $\frac{C_1}{C_2} = \frac{V_2}{V_1}$

    So we need to find how the applied voltage divides across the two capacitors.

    Looking at the network (a typical ladder with resistors and capacitors), in steady state the resistors form a voltage divider. Let me work through the standard configuration where:

    • A voltage source is applied across the network
    • $C_1$ is in series with resistor $R_1$
    • $C_2$ is in series with resistor $R_2$
    • The resistors themselves are in series (since no current flows through the capacitor branches in steady state)

    Step-by-step analysis:

    1. Identify the resistor chain. In steady state, trace the current path through resistors only. Typically this gives us resistors in a series combination, say $3R$ and $5R$ (or similar values that appear in the network).

    2. Apply voltage division. If the total resistance is $3R + 5R = 8R$, and the applied voltage is $V$, then:

       • Voltage across the $3R$ section: $V_1 = V \cdot \frac{3R}{8R} = \frac{3V}{8}$ …

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