Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

An inductor of reactance 100Omega, a capacitor of reactance 50Omega, and a resistor of resistance 50Omega are connected in series with an AC source of 10mathrm~V, 50mathrm~Hz. Average power dissipated by the circuit is ______ W. [cite: 185, 186]

Numerical Answer Type:
Enter a numerical value Answer: 1 to 1 +4 marks

Solution & Explanation

### Related Formula Z = sqrtR^2 + (X_L - X_C)^2 [cite: 786] P = I_textrms^2 R = fracV_textrms^2 RZ^2 [cite: 785, 786] ### Core Logic First, find the total impedance Z of the LCR circuit: [cite: 185, 786] Z = sqrt50^2 + (100 - 50)^2 = sqrt50^2 + 50^2 = 50sqrt2\ Omega [cite: 185, 786] Now calculate the average power dissipation: [cite: 185, 786] P = frac(10)^2 times 50(50sqrt2)^2 = frac100 times 502500 times 2 = frac50005000 = 1\ textW [cite: 787] ### Pattern Recognition Remember that only the resistive element dissipates real thermal power over a complete cycle[cite: 784, 785]. Reactive elements like ideal inductors and capacitors store and release energy alternately without net consumption. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

Reference Study Guides

Alternating Current

More Alternating Current Previous-Year Questions — Page 2

Q16 2025 RMS Value of Alternating Current
An alternating current is given by I=I_Atext sin omega t +I_Btext cos omega t. The r.m.s. current will be :-
  • A. sqrtI_A^2+I_B^2
  • B. fracsqrtI_A^2+I_B^22
  • C. sqrtfracI_A^2+I_B^22
  • D. frac|I_A+I_B|sqrt2

Solution

### Related Formula The root-mean-square current value I_textrms for a periodic function is defined as: I_textrms = sqrtfrac1Tint_0^T I^2 dt For a single sinusoidal term I = I_0sin(omega t + phi), the root-mean-square value simplifies directly to: I_textrms = fracI_0sqrt2 ### Core Logic We can combine the orthogonal sine and cosine components into a single phase-shifted wave: I = I_Asinomega t + I_Bcosomega t = sqrtI_A^2 + I_B^2sin(omega t + phi) where peak current amplitude corresponds to: I_0 = sqrtI_A^2 + I_B^2 ### Step 1: Calculating RMS Value Using the standard peak-to-RMS conversion factor: I_textrms = fracI_0sqrt2 = fracsqrtI_A^2 + I_B^2sqrt2 = sqrtfracI_A^2 + I_B^22 ### Pattern Recognition Orthogonal sine and cosine functions are independent. Their mean squared averages add linearly, so you can think of it as a vector addition: fracI_A^22 + fracI_B^22 under the square root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current
Q1 2025 Choke Coil and AC Circuits
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor left(mathrmR / sqrtmathrmR^2 + omega^2mathrmL^2right) , where omega is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth (A) and (R) are true but (R) is not the correct explanation of (A).
  • B. text(A) is false but (R) is true.
  • C. textBoth (A) and (R) are true and (R) is the correct explanation of (A).
  • D. text(A) is true but (R) is false.

Solution

### Related Formula Z = sqrtR^2 + (omega L)^2 cos phi = fracRZ ### Core Logic A choke coil has a high inductance L and low resistance R. It reduces the current through the mercury tube without wasting electrical power as heat. Statement (A) is correct. The average power dissipation is controlled by the low power factor cos phi. Reason (R) correctly explains that without a choke coil, the direct supply voltage could cause an overflow of current, destroying the tube filament. ### Pattern Recognition Choke coils use high L and low R to minimize power loss while reducing circuit current efficiently. ### Chapter Mix Class 12 Physics: Alternating Current

More Alternating Current Questions — jee_main_2025_07_april_evening

Practice all Alternating Current previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...