Chapter 18, Problem 69QAP

To determine

The value of R₁, R₂, and R₃ in circuit.

Answer to Problem 69QAP

R₁, R₂, and R₃ are 27-ohm, 108 ohm and 108 ohm.

Explanation of Solution

Given:

Voltage present in circuit, V_b= 9 V

Power dissipated in each resistor, P₁= P₂ = P₃ = 1.50 W

Formula used: Power dissipates across resistor

  $P=\frac{V^2}{R}=I^{2}R$.

Where, P is power dissipated

V = Voltage across resistor

I = Current flowing through resistor

R = Resistance..

Equivalent resistance, when resistance connects in series

  $R_{eq}=R_a+R_b$.

Equivalent resistance, when resistance connects in parallel

  $\frac{1}{R_{eq}}=\frac{1}{R_a}+\frac{1}{R_b}$.

Equivalent power, when resistance connects in series

  $\frac{1}{P_{eq}}=\frac{1}{P_a}+\frac{1}{P_b}$.

Equivalent power, when resistance connects in parallel

  $P_{eq}=P_a+P_b$.

Figure:

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Calculation:

In figure (1)

Power in each resistance is equal..

  $\begin{array}{l}{P}_2=P_3\\ \frac{{\left(V_{be}\right)}^2}{R_2}=\frac{{\left(V_{cd}\right)}^2}{R_3}\\ \because bothareconnectinparallel,V_{be}=V_{cd}\\ R_2=R_3\mathrm{......}(a)\end{array}$.

Equivalent Power of R₂and R₃

  $\begin{array}{l}{\left(P_{23}\right)}_{eq}=P_2+P_3\\ {\left(P_{23}\right)}_{eq}=1.5W+1.5W\\ {\left(P_{23}\right)}_{eq}=3W\mathrm{......}(b)\end{array}$.

Equivalent resistance of R₂and R₃

  $\begin{array}{l}\frac{1}{{\left(R_{23}\right)}_{eq}}=\frac{1}{R_2}+\frac{1}{R_3}\\ \because R_2=R_3\\ {\left(R_{23}\right)}_{eq}=\frac{R_2}{2}\mathrm{......}(c)\end{array}$.

In figure (2)

Current flowing through resistance$R_{1}and{\left(R_{eq}\right)}_{23}$is I₁ Power of both resistance$R_{1}and{\left(R_{eq}\right)}_{23}$.

  $\begin{array}{l}{I}_1^2{R}_1=P_1\\ \Rightarrow I_1^2{R}_1=1.5W\mathrm{.......}(d)\\ \text{and,}{I}_1^2{\left(R_{eq}\right)}_{23}={\left(P_{23}\right)}_{eq}\\ \Rightarrow I_1^2\left(\frac{R_2}{2}\right)=3W\mathrm{.......}(e)\end{array}$.

On comparing equation (d) and (e)

  $R_2=4R_1\mathrm{......}(f)$.

So, equivalent resistance of circuit

  $\begin{array}{l}{R}_{eq}=R_1+{\left(R_{23}\right)}_{eq}\\ R_{eq}=R_1+\frac{R_2}{2}\\ \because R_2=4R_1\\ R_{eq}=R_1+\frac{4R_1}{2}\\ R_{eq}=3R_1\mathrm{......}(g)\end{array}$.

Equivalent power of circuit

  $\begin{array}{l}\frac{1}{P_{eq}}=\frac{1}{P_1}+\frac{1}{{\left(P_{23}\right)}_{eq}}\\ \frac{1}{P_{eq}}=\frac{1}{1.5W}+\frac{1}{3W}\\ P_{eq}=1W\mathrm{.......}(h)\end{array}$.

In figure (3)

  $\begin{array}{l}{P}_{eq}=\frac{{\left(V_{eq}\right)}^2}{R_{eq}}\\ \because R_{eq}=3R_1\\ 1W=\frac{{\left(9V\right)}^2}{3R_1}\\ R_1=27\Omega \end{array}$

From equation (f),

  $R_2=4R_1=108\Omega$.

From equation (a),

  $R_3=R_2=108\Omega$.

Conclusion:

Thus, the value of$R_1=27\Omega \text{and}{R}_2=R_3=108\Omega$.