Compare the power used in the $2 \Omega$ resistor in each of the following circuits:
$(i)$ a $6\ V$ battery in series with $1\ \Omega$ and $2\ \Omega$ resistors, and
$(ii)$ a $4\ V$ battery in parallel with $12\ \Omega$ and $2\ \Omega$ resistors.

Given: A resistor of $2 \Omega$.

To do: To compare the power used in:

$(i)$ a $6\ V$ battery in series with $1\ \Omega$ and $2\ \Omega$ resistors, and

$(ii)$ a $4\ V$ battery in parallel with $12\ \Omega$ and $2\ \Omega$ resistors.

Solution:

$(i)$. Here, the potential difference $V=6\ V$

$1\ \Omega$ and $2\ \Omega$ resistors are connected in series. So the equivalent resistance $R=R_1+R_2=1\ \Omega+2\ \Omega=3\ \Omega$

So, the current flowing $I=\frac{V}{R}$

$=\frac{6\ V}{3\ \Omega}$

$=2\ A$

So, the power used in $2\ \Omega$ resistor $P=I^2R$

$=2^2\times 2$

$=8\ Watt$

(ii). Here, the potential difference $=4\ V$

Because the $12\ \Omega$ and $2\ \Omega$ resistors are connected in parallel, so the potential difference will be the same.

So, power consumed in $2\ \Omega$ will be:

$P=\frac{V^2}{R}$

Or $P=\frac{4^2}{2}$

Or $P=\frac{16}{2}$

Or $P=8\ Watt$

Now, on comparing both $(i)$ and $(ii)$ circuits, we find that the power used in the $2\ \Omega$ resistor is the same.

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Updated on: 10-Oct-2022

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