Four resistances of 10 Ω , 60 Ω , 100 Ω and 200 Ω respectively taken in order are used to form a square loop. A 15 V battery is connected across the ends of the 200 Ω resistance, the current through i

Question

Four resistances of 10 Ω , 60 Ω , 100 Ω and 200 Ω respectively taken in order are used to form a square loop. A 15 V battery is connected across the ends of the 200 Ω resistance, the current through it will be

TestHub · Accepted Answer

Here, the resistance for 10 Ω , 60 Ω and 100 Ω are in series and they together are in parallel to 200 Ω resistance when a potential difference of 15V is applied across 200 Ω , then current through it is I = 15 200 = 7.5 × 10 - 2 A

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