JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

  • question_answer
    If \[{{\theta }_{i}}\] is the inversion temperature, \[{{\theta }_{n}}\] is the neutral temperature, \[{{\theta }_{c}}\] is the temperature of the cold junction, then                   [AIEEE 2002]

    A) \[{{\theta }_{i}}+{{\theta }_{c}}={{\theta }_{n}}\]

    B) \[{{\theta }_{i}}-{{\theta }_{c}}=2{{\theta }_{n}}\]

    C) \[\frac{{{\theta }_{i}}+{{\theta }_{c}}}{2}={{\theta }_{n}}\]

    D) \[{{\theta }_{c}}-{{\theta }_{i}}=2{{\theta }_{n}}\]

    Correct Answer: C

    Solution :

    [c] In Seebeck (thermoelectric) effect, the temperature of hot junction at which the thermo-emf is maximum is called the neutral temperature \[({{\theta }_{n}})\] and the temperature at which thermo-emf changes its sign is called inversion temperature \[({{\theta }_{i}})\]. If \[({{\theta }_{c}})\] is the temperature of cold junction, then these three are related by the expression             \[\therefore \] \[{{\theta }_{n}}-{{\theta }_{c}}={{\theta }_{i}}-{{\theta }_{n}}\]             \[{{\theta }_{n}}=\frac{{{\theta }_{i}}+{{\theta }_{c}}}{2}\]


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