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

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0.67 A

0.27 A

0.47 A

0.57 A

answer is A.

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Detailed Solution

0.67 A

current would flow through the

12 ohm

resistor.
Given,
Battery

= 9 V

Resistors of

0.2 ohm, 0.3 ohm, 0.4 ohm, 0.5 ohm and 12 ohm

are connected in series to the battery.

R_{1} = 0.2 ohm

R_{2} = 0.4 ohm

R_{3} = 0.3 ohm

R_{4} = 0.5 ohm

R_{5} = 12 ohm

As all are connected in a series effective resistance would be :

R_{eff} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}

= 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 ohm

By ohm's law:

V = I \times R

I = \frac{V}{R}

\frac{9}{13.4}

= 0.67 A

As resistors are connected in series.

I = I_{1} = I_{2} = I_{3} = I_{4} = I_{5}

Hence, current across

12 ohm

the resistor is

0.67 A

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