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A resistance of 4 Omega and a wire of le...

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

`1.5V`

B

`3.0V`

C

`0.67V`

D

`1.33V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E=pil=(Vl)/L=(iR)/LxxIimpliesE=E/(R+R_(h)+r) xxR/L xxl`
`implies E=10/(5+4+1) xx5/5xx3=3V`
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