Asked by Anonymous
A 20v battery with an internal resistance of 5ohm is connected to a resistance of x ohm if an additional 6ohm resistance is connected across the battery find the value of x so that the power supplied externally by the battery remains the same
Answers
Answered by
Henry
Io = 20/(5+6) = 1.82A. = Output current.
Po = (1.82)^2 * 6 = 19.83 Watts.
Io^2 * Rt = 19.83.
(1.82)^2 * Rt = 19.83,
Rt = 2.06 Ohms.
6*X/(6+X) = 2.06.
6X = 12.36 + 2.06X,
3.94X = 12.36, X = 3.14 Ohms.
Po = (1.82)^2 * 6 = 19.83 Watts.
Io^2 * Rt = 19.83.
(1.82)^2 * Rt = 19.83,
Rt = 2.06 Ohms.
6*X/(6+X) = 2.06.
6X = 12.36 + 2.06X,
3.94X = 12.36, X = 3.14 Ohms.
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