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`AB` and `CD` are two uniform resistance wires of lengths `100 cm` and `80 cm` respectively. The connections are shown in the figure. The cell of emf `5 V` is ideal while the other cell of `E` has internal resistance `2 Omega`. A length of `20 cm` of wire `CD` is balanced by `40 cm` of wire `AB`. Find the emf `E` in volt, if the reading of the ideal ammeter is `2 A`. The other connecting wires have negligible resistance |
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Answer» Correct Answer - 12 Potential difference across wire AB = 5V `:.` p.d across 40 cm of this wire `= (5)/(100) xx 40` = 2 volt `:.` Potential difference across 20 cm of wire CD = 2 volt `:.` p.d across wire `CD =(2)/(20)xx80=8` volt p.d across `2 Omega` resistance `= 2 xx 2 = 4` volt `:.` Emf of the cell = 12 volt. |
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