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.