A battery of 9 V is connected in series with resistors of $0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω$ and $12 Ω$ respectively. How much current will flow through the $12 Ω$ resistor?

Question

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Answer 1

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

Step 1 Finding the equivalence resistance
For resistors in series,
$R_{e q} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}$
$= 0.2 + 0.3 + 0.4 + 0.5 + 12$
$= 13.4 Ω$

Step 2: Finding the current.
Using the ohm's Law:
$V = I R_{e q}$
$9 = 13.4 I$
$I = 0.67 A$

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

Answer 2

Hint:

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

Step 1 Finding the equivalence resistance

For resistors in series,

R_{e q} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}

= 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 Ω

Step 2: Finding the current.

Using the ohm's Law:

V = I R_{e q}

9 = 13.4 I

I = 0.67 A

When resistors are connected in series, the current is same in all the resistors. Hence, current in

12 Ω

resistor

= 0.67 A

.