the four arms of wheatstone network are of resistances 5 ohm, 2 ohm, 6 ohm and 4ohm respectively. one terminal of a galvanometer if resistance 20 ohm is connected to the junction of 6 ohm and 4 ohm. the other two terminals are connected to a cell of emf 2v and negligible resistance. find the current through the galvanometer
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Answer:
0.082 A
Explanation:
R bridge = (5 + 2) × (6+ 4) / 7 + 10
= 7 × 10/ 17
= 70/ 17
R net = 70/ 17 + 20 = (70 + 340)/17 = 410/ 17
Current I = 2 × 17/ 410 = 34/ 410 = 0.082 A
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