Question
Download Solution PDFThree resistors of 5 Ω are connected in series, and a battery of 10 V is connected to the two end of the series combination of resistors. Calculate the current drawn from the battery.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is 0.67A.
Given:
Three resistors of 5 Ω are connected in series and \(V=10V\)
Formula:
Ohm's law: \(V=IR\)
Calculations:
As the three resistors are connected in series, their equivalent resistance is:
\(Rs=5+5+5=15Ω\)
Now by applying Ohm's law we get:
\(V=IR\)
\(I={V \over R}={10 \over 15}=0.67A\)
Thus, the current flowing is 0.67A.
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