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In the given circuit for ideal diode, the current through the battery is

Option: 1

0.5


Option: 2

1.5


Option: 3

1.0


Option: 4

2


Answers (1)

best_answer

The current through the battery is
\mathrm{I=\frac{10 \mathrm{~V}}{5 \Omega+5 \Omega}=\frac{10 \mathrm{~V}}{10 \Omega}=1 \mathrm{~A} }

Posted by

Nehul

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