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Solution

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Use the Ohm's law and the equivalence resistance in series combination.

For resistors in series,

**Step 2: Finding the current.**

$R_{eq}=R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$

$=0.2+0.3+0.4+0.5+12$

$=13.4Ω$

Using the ohm's Law:

$V=IR_{eq}$

$9=13.4I$

$I=0.67A$

When resistors are connected in series, the current is same in all the resistors. Hence, current in $12Ω$ resistor $=0.67A$.

Video Explanation

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