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A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).
(Round off up to 2 decimal places)
    Correct answer is '0.29'. Can you explain this answer?
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    A part of circuit in a steady state along with the currents flowing i...
    Using Kirchhoff's first law at junction a and b, we have found the current in other wires of the circuit on which currents were not shown.
    Now, to calculate the energy stored in the capacitor, we will have to first find the potential difference Vab across it.
    ∴ Va - 3 x 5 - 3 x 1 + 3 x 2 = Vb
    ∴ Va - Vb = Vab = 12 V
    = ½(4 x 10-6)(12)2 J = 0.29 mJ
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    A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).(Round off up to 2 decimal places)Correct answer is '0.29'. Can you explain this answer?
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    A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).(Round off up to 2 decimal places)Correct answer is '0.29'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).(Round off up to 2 decimal places)Correct answer is '0.29'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances, etc. is shown in the figure. Calculate the energy (in mJ) stored in the capacitor C (4μF).(Round off up to 2 decimal places)Correct answer is '0.29'. Can you explain this answer?.
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