Chapter 27: Q86P (page 801)

Question: Two resistors $R_{1} a n d R_{2}$ may be connected either in series or in parallel across an ideal battery with emf $蔚$ . We desire the rate of energy dissipation of the parallel combination to be five times that of the series combination. Iflocalid="1662972116950" $R_{1} = 100 惟$ , what are the (a) smaller and (b) larger of the two values of $R_{2}$ that result in that dissipation rate?

Short Answer

Expert verified

a)The smaller value of $R_{2}$ that results in the dissipation rate is $38 惟$ $38 惟$ .
b)The larger value of $R_{2}$ that results in the dissipation rate is $262 惟$ .

Step by step solution

The given data

a)The rate of dissipation energy of the parallel combination is desired to be five times of the series combination, that is $\frac{P_{p}}{P_{s}} = 5$ .
b) Value of resistance of resistor 1, $R_{1} = 100 惟$

Understanding the concept of dissipation rate

The rate of dissipation energy is the energy rate that is generated due to the charging and discharging of the components of the circuit. Thus, using the equivalent resistance condition of both series and parallel conditions, the required possible values of resistance can be given.

Formulae:

The equivalent resistance for a series combination,

$R_{e q} = {鈭�}_{1}^{n} R_{I}$ $R_{e q} = {鈭� R_{I}}_{1}^{n}$ (i)

The equivalent resistance for a parallel combination,

$R_{e q} = {鈭�}_{1}^{n} \frac{1}{R_{I}}$ (ii)

The rate of dissipation energy of a given combination,

$P = \frac{蔚^{2}}{R_{e q}}$ (iii)

Here $蔚$ is the emf of the battery.

The roots of a quadratic equation $a x^{2} + b x + c = 0$ can be calculated as,
$x = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a} (i v)$

a) Calculation of the smaller resistance value

Now, for the combination of series resistors, the rate of dissipation energy for this combination is given using equation (i) in equation (iii) as follows:

$P_{S} = \frac{蔚^{2}}{R_{1} + R_{2}} 蔚^{2} = P_{s} (R_{1} + R_{2})$ (a)

Similarly, the rate of dissipation energy for the parallel combination of resistors can be given using equation (ii) in equation (iii) as follows:

$P_{p} = \frac{蔚^{2}}{1 I (\frac{1}{R_{1}} + \frac{1}{R_{2}})} = \frac{蔚^{2}}{1 I (\frac{R_{1} + R_{2}}{R_{1} R_{2}})} P_{p} = \frac{蔚^{2} (R_{1} + R_{2})}{R_{1} R_{2}}$

Substituting values from the equation a and we get,

$P_{p} = \frac{P_{s} {(R_{1} + R_{2})}^{2}}{R_{1} R_{2}} \frac{P_{p}}{P_{s}} = \frac{{(R_{1} + R_{2})}^{2}}{R_{1} R_{2}} \frac{{(R_{1} + R_{2})}^{2}}{R_{1} R_{2}} = 5 R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} = 5 R_{1} R_{2} R_{1}^{2} + R_{2}^{2} = 3 R_{1} R_{2}$ $P_{p} = \frac{P_{s} {(R_{1} + R_{2})}^{2}}{R_{1} R_{2}} \frac{P_{p}}{P_{s}} = \frac{{(R_{1} + R_{2})}^{2}}{R_{1} R_{2}} \frac{{(R_{1} + R_{2})}^{2}}{R_{1} R_{2}} = 5 R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} = 5 R_{1} R_{2} R_{1}^{2} + R_{2}^{2} = 3 R_{1} R_{2}$

Substituting values in the above equation, and we get,

${(100 惟)}^{2} + R_{2}^{2} = 3 (100 惟) R_{2} R_{2}^{2} - (300 R_{2}) 惟 + 10000 惟^{2} = 0$

The above equation is a quadratic equation whose roots value can be given using equation (iv) as follows:

(a=1,b=-300,c=10000)
localid="1662975514013" $R_{2} = \frac{- (- 300) 卤 \sqrt{{(- 300)}^{2} - 4 (1) (10000)}}{2 (1)} = \frac{300 卤 \sqrt{90000 - 40000}}{2} = \frac{300 - 224}{2} o r \frac{300 + 224}{2} = 38 惟 o r 262 惟$