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The total current supplied to the circuit by the battery is [Figure: a circular wire with a centre node O. The circle is broken by a 6 V battery (upper left) and a 3 Ω resistor (right). The upper arc between them, node X, receives the outer ends of a 2 Ω and a 6 Ω radial resistor. The lower arc, node Y, receives the outer end of a 1.5 Ω radial resistor. The inner ends of all three radial resistors meet at O. So the battery and the 3 Ω are both connected between X and Y.]

Asked in JEE Main 2004 · Series and parallel combinations

• (1) 1 A
• (2) 2 A
• (3) 4 A
• (4) 6 A

Step-by-step solution

Why the other options are wrong

• (1) No standard slip gives 1 A; it would need 6 Ω total.
• (2) Uses only the 3 Ω branch: 6/3=2 A.
• (4) No standard slip gives 6 A; it would need 1 Ω total.

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