Question

Rank in order, from largest to smallest, the powers $P_{\mathrm{a}}$to $P_{\mathrm{d}}$dissipated by the four resistors in FIGURE

Short answer

The rank in which power is dissipated, from largest to smallest, is$P_c>P_d>P_a>P_b$

Step-by-step solution

. Given information

Given :

Voltage$V_a=\Delta V$

Resistance$R_a=R$

Power $P_a=?$

Voltage$V_b=0.5\Delta V$

Resistance, $R_b=2R$

Power $P_b=?$

Voltage $V_c=2\Delta V$

Resistance$R_c=0.5R$

Power $P_c=?$

Voltage $V_d=2\Delta V$

Resistance$R_d=2R$

Power$P_d=?$

Step 2. 

The potential difference across the first resistor is $\Delta V$, and the resistance across the first resistor is R, as shown in figure (a).

The first resistor's power dissipated is,

$P_a=\frac{(\Delta V{)}^2}{R}$The potential difference across the second resistor in the figure (b) is $\frac{1}{2}\Delta V$, and the resistance across the second resistor is $2R$.

The second resistor's power dissipated is,

$\begin{array}{rcl}{P}_b& =& \frac{{\left(\frac{1}{2}\Delta V\right)}^2}{2R}\\ & =& \frac{1}{8}\frac{(\Delta V{)}^2}{R}\end{array}$

The potential difference across the third resistor in figure (c) is $2\Delta V$, and the resistance across the third resistor is$\frac{1}{2}R$.

The third resistor's power dissipated is,

$\begin{array}{rcl}{P}_c& =& \frac{(2\Delta V{)}^2}{\frac{1}{2}R}\\ & =& 8\frac{(\Delta V{)}^2}{R}\end{array}$

The potential difference across the fourth resistor in figure (d) is$2\Delta V$, and the resistance across the fourth resistor is $2R$.

The fourth resistor's power wasted is,

$\begin{array}{rcl}{P}_d& =& \frac{(2\Delta V{)}^2}{2R}\\ & =& 2\frac{(\Delta V{)}^2}{R}\end{array}$

The rank of power dissipated by the four resistors, from greatest to smallest, is $P_c>P_d>P_a>P_b$ when the values of $P_a,P_b,P_c$ and $P_d$ are compared.