Question

In the circuit of the figure, for what value of R will the ideal battery transfer energy to the resistors at a rate of 57.0 W? (Note, if you want an open circuit or R = infinity, put R = 1.0×10⁶ Ω,) R1 = 9.0 Ω, R2 = 4.0 Ω and R3 = 4.00 Ω.

Answer 1

Given that, Resistances, \(\mathrm{R}_{1}=9 \Omega\)

$$ \begin{array}{l} \quad \mathrm{R}_{2}=4.00 \Omega \\ \text { Power, } \mathrm{P}=57.0 \mathrm{~W} \\ \text { Voltage, } \mathrm{V}=24.0 \mathrm{~V} \end{array} $$

Since, power \(P=V \mid\)

$$ I=P / V=2.375 \mathrm{~A} $$

The net resistance in the circuit is \(, \mathrm{R}_{\mathrm{s}}=\mathrm{V} / \mathrm{I}\)

$$ \begin{array}{l} =24 / 2.375 \\ =10.105 \Omega \end{array} $$

In the circuit, \(\mathrm{R}_{2}=4.00 \Omega, \mathrm{R}, 4.00 \Omega\) are in parallel. Resultant resistance is, \(\left(1 / \mathrm{R}_{\mathrm{p}}\right)=(1 / 4)+(1 / 4)+(1 / \mathrm{R})\)

$$ \begin{aligned} \text { net resistance, } R_{s}=R_{1}+R_{p}=4 R /(2 R+4) \\ 10.105 \Omega=(9 \Omega)+(4 R /(2 R+4)) \\ 10.105 \Omega=(22 R+36) /(2 R+4) \\ R=2.467 \Omega \end{aligned} $$