For the circuit of Fig. 2.65(a).

(a) Determine $R$ such that the source delivers a power of $4 \mathrm{~mW}$ .

(b) Determine $R$ that results in $15 \mathrm{k} \Omega$ resistor to absorb $1.6 \mathrm{~mW}$ of power

(c) Replace $R$ by a voltage source that will cause no power to be absorbed by any resistor.

Question

For the circuit of Fig. 2.65(a).

(a) Determine [latex]R[/latex] such that the source delivers a power of [latex]4 \mathrm{~mW}[/latex].

(b) Determine [latex]R[/latex] that results in [latex]15 \mathrm{k} \Omega[/latex] resistor to absorb [latex]1.6 \mathrm{~mW}[/latex] of power

(c) Replace [latex]R[/latex] by a voltage source that will cause no power to be absorbed by any resistor.

Accepted Answer

(a) Total series resistance, [latex]R_{t}=10+15+R=25+[/latex] [latex]R \mathrm{k} \Omega[/latex].

Power delivered by the source

[latex]P=\frac{V^{2}}{R_{t}}=\frac{(12)^{2}}{25+R}=4[/latex]

Solving we get [latex]R=11 \mathrm{k} \Omega[/latex]

(b) [latex]\mathrm{P}(10 \mathrm{k} \Omega)=10 I^{2}=1.6 \mathrm{~mW}, I=0.4 \mathrm{~mA}[/latex]

[latex]\frac{12}{10+R+15}=0.4 \mathrm{~mA} ; R=5 \mathrm{k} \Omega[/latex]

(c) No power absorbed means [latex]I=0[/latex]. Therefore voltage source is [latex]12 \mathrm{~V}[/latex] in opposition as in Fig. 2.65(b).

Question 2.ASP.35b: For the circuit of Fig. 2.65(a). (a) Determine R such that t...