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A resistance of `4 Omega` and a wire of length 5 meters and resistance `5 Omega` are joined in series and connected to a cell of e.m.f. `10 V` and internal resistance `1 Omega`. A parallel combination of two identical cells is balanced across `300 cm` of wire. The e.m.f. `E` of each cell is A. 1.5 VB. 3.0 VC. 0.67 VD. 1.33 V |
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Answer» Correct Answer - B `E=pil=(Vl)/(L)=(iR)/(L)xxl` `impliesE=(E)/(R+R_(h)+r)xx(R)/(L)xxl` `impliesR=(10)/(5+4+1)xx(5)/(5)xx3=3V` |
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