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In the circuit shown below `E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega` and `R_(3) = 2 Omega`. The current `I_(1)` is A. `1.6A`B. `1.8A`C. `1.25A`D. `1.0A` |
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Answer» Correct Answer - B |
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