Asked by Anonymous

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?
Answered by Henry
R + 11 = (11A/6.6A)R 1.666R
1.666R - 1R = 11
0.666R = 11
R = 16.5 Ohms. = Original Resistance.

Answered by Henry
Correction: R + 11=(11A/6.6A)*R= 1.666R.

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