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The current in a series circuit is 11.0 A. When an additional 11.0-Ω resistor is inserted in series, the current drops to 6.6 A. What is the resistance in the original circuit?

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E/R = 11.0
E/(R+11) = 6

Since the voltage drop is the same in both cases,

11.0R = 6.6(R+11)
R = 16.5Ω

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