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A battery having 12V emf and internal resistance `3Omega` is connected to a resistor. If the current in the circuit is 1A, then the resistance of resistor and lost voltage of the battery when circuit is closed will beA. `7Omega,7V`B. `8Omega,8V`C. `9Omega,9V`D. `9Omega,10V` |
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Answer» Correct Answer - C Here, `epsi=12V, r=3Omega, I=1A,V=IR=epsi-Ir` `therefore R=(epsi-Ir)/(I)=(12-1xx3)/(1)=12-3=9Omega` and `V=IR=1xx9=9V` |
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