Q. 6 Rank in order, from largest to s... [FREE SOLUTION]

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Chapter 28: Q. 6 (page 789)

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

Short Answer

Expert verified

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 $V_{a} = 螖 V$

Resistance $R_{a} = R$

Power $P_{a} = ?$

Voltage $V_{b} = 0.5 螖 V$

Resistance, $R_{b} = 2 R$

Power $P_{b} = ?$

Voltage $V_{c} = 2 螖 V$

Resistance $R_{c} = 0.5 R$

Power $P_{c} = ?$

Voltage $V_{d} = 2 螖 V$

Resistance $R_{d} = 2 R$

Power $P_{d} = ?$

Step 2.

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

The first resistor's power dissipated is,

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

The second resistor's power dissipated is,

$\begin{array}{rcl} P_{b} & = & \frac{{(\frac{1}{2} 螖 V)}^{2}}{2 R} \\ = & \frac{1}{8} \frac{(螖 V)^{2}}{R} \end{array}$

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

The third resistor's power dissipated is,

$\begin{array}{rcl} P_{c} & = & \frac{(2 螖 V)^{2}}{\frac{1}{2} R} \\ = & 8 \frac{(螖 V)^{2}}{R} \end{array}$

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

The fourth resistor's power wasted is,

$\begin{array}{rcl} P_{d} & = & \frac{(2 螖 V)^{2}}{2 R} \\ = & 2 \frac{(螖 V)^{2}}{R} \end{array}$

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.