Question

Two identical resistors, each of resistance 2 Ohm, are connected in torch
(i) in series and
(ii) in parallel, to a battery of 12 volts. Calculate the ratio of power consumed in two cases.

Answer 1

Hint: In the given question, when the two resistors are connected in series same current flows through each resistor. When the two resistors are connected in parallel across the battery, they have equal potential difference across them. The power consumed by the torch is the product of the potential difference across the resistors and the current flowing through them.

Complete step by step answer:
Given;
The two resistors \[{R_1} = {R_2} = 2\Omega \].
The potential of the battery, \[V = 20\;{\rm{V}}\].
(i) Considering the first case in which the two resistors are connected in series, the resultant resistance of the connection is given as,
\[R = {R_1} + {R_2}\]
On substituting the values in the above expression, we get,
\[\begin{array}{l}
R = \left( {2 + 2} \right)\Omega \\
R = 4\Omega
\end{array}\]
The current that will be flowing through each resistor will be,
\[\begin{array}{l}
I = \dfrac{V}{R}\\
I = \dfrac{{12}}{4}\\
I = 3{\rm{ A}}
\end{array}\]
So, the power consumed in this case is,
\[\begin{array}{l}
{P_1} = V \cdot I\\
{P_1} = 12 \times 3\\
{P_1} = 36{\rm{ W}}
\end{array}\]
Therefore, the power consumed when the resistors are connected in series is 36 W.

(ii) Considering the second case in which the two resistors are connected in parallel, the resultant resistance of the connection is given as,
\[\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\]
On substituting the values in the above expression, we get,
\[\begin{array}{l}
\dfrac{1}{R} = \left( {\dfrac{1}{2} + \dfrac{1}{2}} \right)\Omega \\
\dfrac{1}{R} = 1\Omega
\end{array}\]
The current that will be flowing through each resistor is,
\[\begin{array}{l}
I = \dfrac{V}{R}\\
I = \dfrac{{12}}{1}\\
I = 12{\rm{ A}}
\end{array}\]
So, the power consumed in this case is,
\[\begin{array}{l}
{P_2} = V \cdot I\\
{P_2} = 12 \times 1\\
{P_2} = 12{\rm{ W}}
\end{array}\]
Therefore, the power consumed when the resistors are connected in series is 12 W.
Hence, the ratio of the powers is calculated as,
 \[\begin{array}{l}
\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{36}}{{12}}\\
\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{3}{1}\\
\dfrac{{{P_1}}}{{{P_2}}} = 3:1
\end{array}\]
Therefore, the ratio of the power consumed is \[3:1\].

Note:In this question, we can see that when the resistors are connected in series, they draw more power from the battery than when the resistors are connected in parallel. In a series connection, the battery's potential gets divided according to the resistors' values, while in parallel connection, the potential difference is the same across both the resistors. On the other hand, in series connection, the current flowing through all the resistors is the same while it gets divided according to the resistors' values in parallel connection.