#electrical measurement

AC Fundamentals - RMS, Average and Peak Values

What is the significance of the rms value in specifying AC quantities?

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. The rms (root mean square) value of an AC quantity (e.g., voltage or current) is the equivalent DC value that produces the same average power in a resistive load. It accounts for the time-varying nature of AC, making it a standard measure for power calculations. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 8 \, \Omega \), \( R_2 = 16 \, \Omega \), \( R_3 = 12 \, \Omega \), \( R_4 = 24 \, \Om

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. Check balance: (R₁/R₂) = (8/16) = 0.5 , (R₃/R₄) = (12/24) = 0.5 . Bridge is balanced. Since balanced, I_g = 0 A . 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 0 A,

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

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Wheatstone Bridge and Meter Bridge

In a Wheatstone bridge, if the bridge is unbalanced, what determines the direction of current through the galvanometer?

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. When unbalanced ( R₁ / R₂ neq R₃ / R₄ ), the potentials at the galvanometer’s nodes differ. The current flows from the higher-potential node to the lower one, determined by the relative voltage drops across the bridge arms. 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

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

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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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