Practice portal › Current Electricity › Combination of Resistors
Asked in GUJCET 2009; GUJCET 2016 · Series and parallel combinations
Given (figure): a 9 V battery across two main branches. Upper branch: 1.5 Ω in series with 2 Ω and 6 Ω in parallel. Lower branch: 3 Ω.
In parallel, 2 Ω and 6 Ω give (2×6)/(2+6) = 1.5 Ω.
Upper branch: 1.5 + 1.5 = 3 Ω.
The upper branch in parallel with the 3 Ω gives R_eq = (3×3)/(3+3) = 1.5 Ω.
Total current: I = 9/(1.5) = 6 A, that is 3 A in each main branch.
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