Question

For what value of $R$ power in $R$ is maximum?

Answer 1

Hint:The concept of power and its formula is applied to determine its value. Kirchhoff's law given by the junction rule for current and the loop rule for voltage is applied in order to determine equations relating the voltage current and resistor values. To determine the maximum power the differentiation is performed.

Complete step by step answer:
The above problem revolves around the concept of power dissipated in a circuit. Let us first know the concept and the definition of power.Power is defined as the rate at which the work is done by a source of emf or voltage supply in maintaining an electric current throughout the circuit. The power is a measure of the heat that is dissipated due to the production of current when there is a voltage supply connected across its circuit elements.

However, in this question we are asked to consider a criteria wherein the power dissipated across $R$ must be maximum. Since, we are asked to find the value of $R$ for which power is maximum we are first required to find the current through the resistor $R$ and then substitute this value in the power equation relating current and resistance. To do so we first need to determine the equation for power in terms of current and resistance. The formula for power is given by the equation:
$P = IV$
We now apply the equation for ohm’s law which gives the relationship between current and voltage values in a circuit:
$V = IR$

This value for $V$ is substituted in the above power equation to get:
$P = {I^2}R$
In order to construct equations relating the voltage, current and resistance values given in the circuit we need to apply Kirchhoff's law. This is done in order to find the current that passes through the resistor $R$. Since, there are junctions present in the circuit the current gets split up to go in different paths in the circuit. This is given by Kirchhoff's junction law. A junction is nothing but a point in an electric circuit where two or more conductor wires are joined together.

The Kirchhoff first law that governs the current in a circuit states that in any electric circuit the algebraic sum of the currents entering a junction is equivalent to the sum of currents leaving that junction. This means that that current is split up at a junction which can be added up to give the total current that had entered the junction. This can be illustrated in the diagram below:

Since we need to find the power with respect to resistor $R$ the current through resistor $R$ must be considered and since the current that flows through $R$ as per the above diagram is ${I_2}$, we get the power equation to be:
$P = I_2^2R$ --------($1$)
Our aim is to find out this value of ${I_2}$ using the equations that will be constructed using Kirchhoff’s laws.As per the Kirchhoff’s current law from the above diagram we get the equation:
$I = {I_1} + {I_2}$ ----------($2$)
In order to construct the required equations relating all the quantities we need to apply Kirchhoff's second law or the loop rule which is known as Kirchhoff’s voltage law. The Kirchhoff’s loop rule states that the algebraic sum of the voltages in any closed circuit or loop of a circuit is equal to the sum of the products of currents and resistances in it. This was determined from the ohm’s law relation.

It follows the principle that the sum of the voltage drops across the resistors present in the loop in terms of its currents and resistances given by ohm’s law will be equivalent to the source voltage provided to the circuit. Thus, as per Kirchhoff’s voltage law we first consider the loop through $ABEF$ to get the equation:
$20I + {I_2}R = V$ --------($3$)
Next we apply the Kirchhoff’s voltage law to loop $BEDF$ to get the equation:
$30{I_1} = {I_2}R$
By rearranging the terms to make ${I_1}$ as the subject we get:
${I_1} = \dfrac{{{I_2}R}}{{30}}$
This value of ${I_1}$ is substituted into the equation ($2$) to get:
$I = \dfrac{{{I_2}R}}{{30}} + {I_2}$
Taking out the common term ${I_2}$ we get:
$I = {I_2}\left[ {\dfrac{R}{{30}} + 1} \right]$
By taking the LCM we get:
$I = {I_2}\left[ {\dfrac{{R + 30}}{{30}}} \right]$

This above value is substituted into equation ($3$) in place of $I$ to get:
$20{I_2}\left[ {\dfrac{{R + 30}}{{30}}} \right] + {I_2}R = V$
By further simplifying this equation we get:
$\dfrac{{20{I_2}R}}{{30}} + 20{I_2} + {I_2}R = V$
$ \Rightarrow \dfrac{2}{3}{I_2}R + 20{I_2} + {I_2}R = V$
We now bring the similar terms together to get:
$\dfrac{2}{3}{I_2}R + {I_2}R + 20{I_2} = V$
By taking out the common terms we get:
${I_2}R\left[ {\dfrac{2}{3} + 1} \right] + 20{I_2} = V$
$ \Rightarrow \dfrac{5}{3}{I_2}R + 20{I_2} = V$
By taking out the common term ${I_2}$ we get:
${I_2}\left[ {\dfrac{5}{3}R + 20} \right] = V$
We now get the equation for ${I_2}$ by making ${I_2}$ as the subject which is given as:
${I_2} = \dfrac{V}{{\left[ {\dfrac{5}{3}R + 20} \right]}}$

Now that we have found out the required current value through resistor $R$ we will now substitute this current in the power equation. Hence, by substituting this value for ${I_2}$in equation ($1$) we get:
$P = {\left( {\dfrac{V}{{\dfrac{5}{3}R + 20}}} \right)^2}R$
On further simplification:
$P = \dfrac{{{V^2}R}}{{{{\left[ {\dfrac{5}{3}R + 20} \right]}^2}}}$ ------------($4$)
In order to find the maximum value for power it must be differentiated and equated to zero so as to maximize the value of power that is dissipated. This is because the peak or the maxima of power value is required and hence the slope or the tangent drawn to this peak value in the graph for power that is dissipated in cycles will be parallel to the x-axis, thus making its derivative or slope value equal to zero.Hence, in order to obtain the maximum power we need to differentiate it and equate it to zero and hence we get:
$\dfrac{{dP}}{{dR}} = 0$

We differentiate the power with respect to $R$ because we are asked to find out the maximum power across resistor $R$ only.We now substitute equation ($4$) in place of $P$ in the above differentiation equation for maximum power we get:
$\dfrac{d}{{dR}}\left( {\dfrac{{{V^2}R}}{{{{\left[ {\dfrac{5}{3}R + 20} \right]}^2}}}} \right) = 0$
We perform the differentiation by applying the quotient rule and the chain rule for differentiation to get:
$\dfrac{{{V^2}{{\left( {\dfrac{5}{3}R + 20} \right)}^2} - 2\left( {\dfrac{5}{3}R + 20} \right) \times \dfrac{5}{3} \times {V^2}R}}{{{{\left( {\dfrac{5}{3}R + 20} \right)}^4}}} = 0$
Taking out the common term $\left( {\dfrac{5}{3}R + 20} \right)$ out we get:
$\dfrac{{\left( {\dfrac{5}{3}R + 20} \right)\left[ {{V^2}\left( {\dfrac{5}{3}R + 20} \right) - 2 \times \dfrac{5}{3} \times {V^2}R} \right]}}{{{{\left( {\dfrac{5}{3}R + 20} \right)}^4}}} = 0$
The common term from the numerator and denominator gets cancelled out to get:
$\dfrac{{\left[ {{V^2}\left( {\dfrac{5}{3}R + 20} \right) - 2 \times \dfrac{5}{3} \times {V^2}R} \right]}}{{{{\left( {\dfrac{5}{3}R + 20} \right)}^3}}} = 0$

We now perform cross multiplication to get:
${V^2}\left( {\dfrac{5}{3}R + 20} \right) - \dfrac{{10}}{3} \times {V^2}R = 0$
$ \Rightarrow {V^2}\left( {\dfrac{5}{3}R + 20} \right) = \dfrac{{10}}{3} \times {V^2}R$
By cancelling out the common term ${V^2}$ we get:
$20 = \dfrac{{10}}{3}R - \dfrac{5}{3}R$
We take the LCM to get:
$20 = \dfrac{{10 - 5}}{3}R$
On further simplification we obtain:
$ \Rightarrow 20 = \dfrac{5}{3}R$
$ \Rightarrow 5R = 60$
$ \therefore R = 12\Omega $

Hence, $12\Omega $ is the resistance value for the circuit with maximum power dissipated across the resistor $R$.

Note:A common error that can be observed in this problem is that the equivalence resistance is often calculated and substituted in the power equation which is incorrect. To maximize the power differentiation is required to be carried out. The mathematics and the concept of differentiating must be known in order to solve these types of problems since chain rule and quotient rule must be applied to solve for the value of $R$.