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Asked in GUJCET 2017 · Kirchhoff's laws in networks
Given (figure): an ideal 12 V battery with 4 Ω and 6 Ω in series; a 12 Ω branch containing S₁ is connected across the 6 Ω; S₂ is connected across the 5 Ω on the right.
S₁ open: no current can flow in the 12 Ω branch, so it drops out.
S₂ closed: a zero-resistance path is in parallel with the 5 Ω, so the 5 Ω is short-circuited and carries no current.
What remains is 4 Ω and 6 Ω in series: R = 10 Ω.
I = (12)/(10) = 1.2 A, the same through the 4 Ω and the 6 Ω.
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