2. Solved Papers
  3. BCECE Medical

BCECE Medical BCECE Medical Solved Papers-2002

  • question_answer
    For driving current of 2 A for 6 min in a circuit, 1000 J of work is to be done. The emf of the source in the circuit is :

    A)  1.38V         

    B)  1.68V

    C)  2.03V         

    D)  3.10V

    Correct Answer: A

    Solution :

    If \[i\] is the current in a wire then the charge flown through the wire in t second is                 \[q=i\,t=2\times 6\times 60=720\,\,C\] The work done in taking q coulomb (720 C) of charge from one end of wire to other end under a potential difference of V volts (in this case emf) is \[\Rightarrow \] \[V=\frac{W}{q}=\frac{1000}{720}=1.38\,\,V\]


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