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 would flow through the 12 Ω resistor?

Given: A battery of 9 V is connected in series with resistors of $0.2\ \Omega$, $0.3\ \Omega$, $0.4\ \Omega$, $0.5\ \Omega$ and $12\ \Omega$, respectively.

To do: 

To find the current flowing through the 12 \Omega resistor.

Solution:

Here given, voltage $V=9\ V$

As given the resistors are connected in series, $R$ will be the sum of resistances of all resistors.

$R=R_1+R_2+R_3+R_4+R_5$

Or $R=0.2\ \Omega+0.3\ \Omega+0.4\ \Omega+0.5\ \Omega+12\ \Omega$

Or $R=13.4\ \Omega$

As the resistors are connected in series, the current flowing through each component will be the same.

Let $I$ be the current flowing.

We know that $I=\frac{V}{R}$

Or $I=\frac{9\ V}{13.4\ \Omega}$

Or $I=0.671\ A$

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Updated on: 10-Oct-2022

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