Question

Five circuits are shown below. All the batteries have the same voltage $ V $ and all resistors have the same resistance $ R $ . In which circuit does the battery deliver the most power?

Answer 1

Hint :Joule's law, which states that the heat emitted in resistance is proportional to the square of the current flowing through the resistance over a given time, is directly related to Ohm's law.

Complete Step By Step Answer:
The difference in charge between two points is known as voltage. The rate at which charge flows is referred to as current. The inclination of a substance to resist the movement of charge is known as resistance (current).
We know that $ Power = V \times I $ where $ V $ represents the applied voltage and $ I $ represents the current in the circuit.
According to Ohm ’s Law, we know that $ \dfrac{V}{I} = R $
We can rewrite it as $ I = \dfrac{V}{R} $
Now we substitute $ I $ in the formula of power. So, we get,
 $ Power = V \times I = V \times \dfrac{V}{R} \\
  \therefore Power = \dfrac{{{V^2}}}{R} \\ $
We have to find which battery delivers the most power, from the above equation it is obvious that power will be the most when resistance is the least.
Considering option $ 1 $ :
Resistance in parallel is,
  $ \dfrac{1}{{{R_P}}} = \left( {\dfrac{1}{R} + \dfrac{1}{{2R}}} \right) \\
   \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{3}{{2R}} \\
   \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{3}{{2R}} \\
   \Rightarrow {R_P} = \dfrac{{2R}}{3} \\ $
 $ {R_S} = R $
Total resistance is, $ {R_T} = {R_P} + {R_S} $
 $ {R_T} = \dfrac{{2R}}{3} + R = \dfrac{{5R}}{3} $
Considering option $ 2 $ :
Resistance in parallel is,
  $ \dfrac{1}{{{R_P}}} = \left( {\dfrac{1}{R} + \dfrac{1}{{2R}}} \right) \\
   \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{3}{{2R}} \\
   \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{3}{{2R}} \\
   \Rightarrow {R_P} = \dfrac{{2R}}{3} \\ $
 $ {R_S} = 0 $
Total resistance is, $ {R_T} = {R_P} + {R_S} $
 $ {R_T} = \dfrac{{2R}}{3} + 0 = \dfrac{{2R}}{3} $
Considering option $ 3 $ :
Resistance in parallel is,
  $ \dfrac{1}{{{R_P}}} = 0 \\
   \Rightarrow {R_P} = 0 \\ $
 $ {R_S} = R + R = 2R $
Total resistance is, $ {R_T} = {R_P} + {R_S} $
 $ {R_T} = 0 + 2R = 2R $
Considering option $ 4 $ :
Resistance in parallel is,
  $ \dfrac{1}{{{R_P}}} = \left( {\dfrac{1}{R} + \dfrac{1}{R}} \right) \\
   \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{2}{R} \\
   \Rightarrow {R_P} = \dfrac{R}{2} \\ $
 $ {R_S} = R $
Total resistance is, $ {R_T} = {R_P} + {R_S} $
 $ {R_T} = \dfrac{R}{2} + R = \dfrac{{3R}}{2} $
Since option $ 2 $ has the least resistance, $ {R_T} = \dfrac{{2R}}{3} $ .
Therefore, option $ 2 $ circuit’s battery delivers the most power.

Note :
In electrical systems, resistance is a critical property that can be calculated. Ohms are the units used to assess resistance. Low-resistance materials allow electricity to flow freely. To allow electricity flow through materials with higher resistance, more voltage (EMF) is required.