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#circuits

19 public questions tagged with this topic.

Wheatstone Bridge and Meter Bridge

A Wheatstone bridge has \( R_1 = 14 \, \Omega \), \( R_2 = 28 \, \Omega \), \( R_3 = 10 \, \Omega \). What is \( R_4 \)

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (14/28) = (10/R₄) . Solve: 0.5 = (10/R₄) ⇒ R₄ = (10/0.5) = 20 Ω . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 20 Ω,

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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PHYSICS

A 25 mH inductor is connected to a 220 V, 50 Hz AC source. What is the rms current?

Given: A 25 mH inductor is connected to a 220 V, 50 Hz AC source. What is the rms current? These values define the system as per NCERT data. Formula: X_L = omega L = 2π f L. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: f = 50 Hz, L = 25 × 10⁻³ H . omega = 2 × 3.14 × 50 = 314 rad/s . X_L = 314 × 0.025 = 7.85 Ω . RMS current: I = V/X_L = 220/7.85 approx 28 A . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A 50 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the peak current?

Given: A 50 μF capacitor is connected to a 220 V, 50 Hz AC source. What is the peak current? Formula: X_C = 1/omega C, omega = 2π × 50 = 314 rad/s. Substitution & Calculation: C = 50 × 10⁻⁶F . X_C = frac1314 × 50 × 10⁻⁶approx 63.7 Ω . RMS current: I = V/X_C = 220/63.7 approx 3.45 A . Peak current: i_m = √2 I = 1.414 × 3.45 approx 4.88 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A 4 Ω resistor dissipates 16 W of power. What is the voltage across it?

Given: A 4 Ω resistor dissipates 16 W of power. What is the voltage across it? These values define the system as per NCERT data. Formula: Power: P = V²/R. This is standard NCERT relation. Substitution & Calculation: Rearrange: V = sqrtP R . Substitute: V = sqrt16 × 4 = sqrt64 = 8 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A 12 V battery with negligible internal resistance is connected to a 3 Ω and 6 Ω resistor in series. What is the power

Given: A 12 V battery with negligible internal resistance is connected to a 3 Ω and 6 Ω resistor in series. What is the power dissipated in the 6 Ω resistor? These values define the system as per NCERT data. Formula: Total resistance: R = 3 + 6 = 9 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Current: I = V/R = 12/9 = 4/3 A . Power: P = I² R = (4/3)² × 6 = 16/9 × 6 = 32/3 approx 10.67 W . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

In a rectifier circuit, the diode conducts when it is:

A diode conducts in forward bias (p-side positive, n-side negative), allowing current to flow through the load, which is the basis of rectification in both half-wave and full-wave circuits.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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Cell Signaling Fundamental PrinciplesI

Signaling pathways are called circuits because they

have nodes and edges with logic, is consistent with established principles of cell signaling, receptor pharmacology and cellular regulation. Experimental measurements of binding parameters, genetic loss-of-function studies and pharmacological interventions all converge on the same interpretation. Related options address neighboring concepts but do not satisfy the precise criterion stated in the question.

Ref: NCERT Biology Class 11–12 Alberts et al Molecular Biology of the Cell Lodish et al, Molecular Cell Biology Cooper & Hausman, The Cell Abbas et al., Cellular and Molecular Immunology (for immunology sections)

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