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For the cubical network of 12 one-ohm resistors driven by a 10 V battery (equivalent resistance 5/6 $\Omega$), the total current supplied by the battery is
A4 A
B8 A
C12 A
D20 A
Answer & Solution
Correct answer: C. 12 A
1. The total current is $3I = \varepsilon / R_{eq}$.
2. Substitute: $3I = 10 / (5/6) = 10 \times (6/5) = 12\ \text{A}$.
3. So the total current is 12 A (and each starting edge carries $I = 4\ \text{A}$).
4. 4 A is the per-edge current, not the total; the others miscompute the ratio.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.18_
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