Question

The electronics supply company where you work has two different resistors, R₁and R₂, in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are R₁ and R₂ in parallel and in series—and you can’t separate these two resistor combinations. You separately connect each resistor network to a battery with emf 48.0 V and negligible internal resistance and measure the power Psupplied by the battery in both cases. For the series combination, P= 48.0 W; for the parallel combination, P= 256 W. You are told that R₁ > R₂. (a) Calculate R₁ and R₂. (b) For the series combination, which resistor consumes more power, or do they consume the same power? Explain. (c) For the parallel combination, which resistor consumes more power, or do they consume the same power?

Short answer

1. The magnitude of R₁and R₂ is$R_1=36.0\Omega \hspace{0.33em}and\hspace{0.33em}{R}_2=12.0\Omega$
2. In series combinationR₁ consumes more power.
3. In parallel combinationR₂ consumes more power.

Step-by-step solution

(a) Determination of the magnitude of the resistances R1 and R2.

The resistances when connected in series, the equivalent resistance is,

$R_{eq}=R_1+R_2$

If the voltage applied is ɛ, then the current in the circuit is,

$I=\frac{\epsilon }{R_1+R_2}$

Thus, the power dissipated is,

$\begin{array}{r}{P}_s=I^2\left(R_1+R_2\right)\\ ={\left[\frac{\epsilon }{R_1+R_2}\right]}^2\left(R_1+R_2\right)\\ =\frac{{\epsilon }^2}{\left(R_1+R_2\right)}\end{array}$

Substitute all the values in the above equation,

$$\begin{gathered} 48.0\hspace{0.33em}W=\frac{(48.0\hspace{0.33em}V{)}^2}{(R_1+R_2)} \\ {R}_1+R_2=48.0\hspace{0.33em}\Omega \end{gathered}$$ ...(i)

The resistances when connected in parallel, the equivalent resistance is,

$R_{eq}=1/R_1+1/R_2$

If the voltage applied is ɛ, then the current in the circuit is,

$I=\frac{\epsilon }{1/R_1+1/R_2}$

Thus, the power dissipated is,

$\begin{array}{r}{P}_p=l_1^2{R}_1+l_2^2{R}_2\\ =\frac{{\epsilon }^2}{R_1^2}{R}_1+\frac{{\epsilon }^2}{R_2^2}{R}_2\\ ={\epsilon }^2\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ ={\epsilon }^2\left(\frac{R_1+R_2}{R_2{R}_1}\right)\\ =256\mathrm{W}\end{array}$

Substitute all the values in the above equation,

$\begin{array}{r}(48.0\mathrm{V}{)}^2\frac{R_1{R}_2}{\left(R_1+R_2\right)}=256\mathrm{W}\\ {\text{Use,}}_1{R}_1+R_2=48.0\mathrm{\Omega }\\ R_1{R}_2=432{\mathrm{\Omega }}^2\end{array}$ ...(ii)

Solve equation (i) and (ii) for both the resistances to get two sets of answers,

$\begin{array}{r}{R}_1=36.0\mathrm{\Omega }\text{and}{R}_2=12.0\mathrm{\Omega }\\ R_1=12.0\mathrm{\Omega }\text{and}{R}_2=36.0\mathrm{\Omega }\end{array}$

Since, it is given in question that R₁ > R₂. Thus the first pair is correct. Thus,$R_1=36.0\Omega \hspace{0.33em}and\hspace{0.33em}{R}_2=12.0\Omega$

(b) Determination of whether the resistor consume more power in series combination.

In the series combination,

The same current flows through the resistances. As the power is given as $\mathrm{P}={\mathrm{I}}^2\mathrm{R}$, so more is the resistance more the power consumed. As established R₁ > R₂ , therefore more power will be lost at R₁.

(c) Determination of whether the resistor consume more power in parallel combination.

In the parallel combination,

The same current does not flow through the resistances but the voltage is same. As the power is given as $\mathrm{P}={\mathrm{V}}^2/\mathrm{R}$, so more is the resistance less is the power consumed. As established R₁ > R₂ , therefore less power will be lost at R₁.