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Question

The emf of a cell E is 15 V as shown in the figure with an internal resistance of 0.5Ω. Then the value of the current drawn from the cell is:
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A
2 A
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B
5 A
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C
1 A
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D
3 A
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Solution

The correct option is C 1 A
7Ω,1Ω,10Ω resistances are connected in series.
Hence an equivalent resistance of 7+1+10=18Ω is connected in parallel with a 6Ω resistance.
Hence, equivalent resistance of 6×186+18Ω=4.5Ω is connected in series with 0.5Ω,8Ω,2Ω.
Hence, net resistance of the circuit is 4.5+0.5+8+2=15Ω
Hence, current drawn from cell=EReq=15V15Ω=1A


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