2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-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                                 \[W=Vq\]             \[\Rightarrow \]   \[V=\frac{W}{q}=\frac{1000}{720}=1.38\,V\]


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