JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    Regarding units of \[C{{R}^{2}}\] (C = capacitance and R = resistance e), study the following statements:
    (i) henry                        
    (ii) \[\frac{\text{volt - second}}{\text{ampere}}\]
    (iii) \[\frac{\text{volt }}{\text{ampere}}\]             
    (iv) \[\frac{\text{joule }}{\text{amper}{{\text{e}}^{2}}}\]
    The correct statements are

    A) (i), (ii) and (iii)

    B) (i), (ii) and (iv)

    C) (iii) and (iv)         

    D) (ii) and (iii)

    Correct Answer: B

    Solution :

    [b] Time constant in \[C-R\]and \[L-R\] circuits are CR and\[\frac{L}{R}\]respectively. Hence          \[CR\equiv L\] or              \[C{{R}^{2}}\equiv L\] i.e., units of \[C{{R}^{2}}\] and L are same. Now   \[|e|=L\left( \frac{di}{dt} \right)\] and \[U=\frac{1}{2}L{{i}^{2}}\] Therefore, units of L or \[C{{R}^{2}}\] are Henry, \[\frac{volt-\sec ond}{ampere}\] and\[joule/amper{{e}^{2}}.\]

You need to login to perform this action.
You will be redirected in 3 sec spinner