A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one quarter of the circumference apart are connected to a 24 V battery, having an internal resistance of 0.5 Ω. What is the current flowing through the battery.
(A) 0.5 A (B) 1A (C) 2A (D) 3A
vi) To find the
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3A current flowing through the battery
Explanation:
A circular loop has a resistance of 40 Ω.
Two points P and Q of the loop, which are one quarter of the circumference apart
Resistance of Each part = (1/4) 40 = 10 Ω & (3/4) 40 = 30 Ω
10 Ω ║ 30 Ω
1/R = 1/10 + 1/30
=> 1/R = (3 + 1)/30
=> R = 30/4
=> R = 7.5 Ω.
internal resistance of 0.5 Ω.
Total Resistance = 7.5 + 0.5 = 8 Ω
Voltage applied = 24V
Current = 24/8 = 3A
3A current flowing through the battery.
option D is correct
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