Answer:
R = 1,746 Ω
Explanation:
The power dissipated in the circuit is
P = V I = V² / R
Let's find the current
R = V² / P
Let's calculate
R = 13²/81
R = 2,096 Ω
This is total resistance
R_total = R + r
R = R_total - r
R = 2,096 -0,350
R = 1,746 Ω
The two values of R when the power dissipated is 81 W are 1.74 ohms or -0.35 ohms.
The given parameters;
The current flowing in the circuit is calculated as follows;
E = IR + Ir
E = I(R + r)
[tex]I = \frac{E}{R+r}[/tex] -----(1)
The power dissipated in the circuit is calculated as follows;
P = I²(R + r)
[tex]I^2 = \frac{P}{R+ r} \\\\I = \sqrt{\frac{P}{R+ r} } \ \ ---(2)[/tex]
solve (1) and (2) together;
[tex]\sqrt{\frac{P}{R+ r} } = \frac{E}{R+r} \\\\\frac{P}{R+ r} = \frac{E^2}{(R+r)^2} \\\\\frac{P}{R+ r} = \frac{E^2}{R^2 + 2Rr + r^2} \\\\E^2(R+r) = P(R^2 + 2Rr + r^2)\\\\13^2(R+ 0.35) = 81(R^2 + 2(0.35)R + (0.35)^2)\\\\169R + 59.15 = 81R^2 + 56.7R + 9.923\\\\81R^2 -112.3R -49.23 = 0\\\\Solve \ the \ quadratic \ equation\ using formula \ method;\\\\a = 81, \ b = -112.3, \ c = -49.23\\\\ R = \frac{-b \ \ +/- \ \ \sqrt{b^2 -4ac} }{2a} \\\\[/tex]
[tex]R = \frac{-(-112.3)\ \ +/- \ \ \sqrt{(-112.3)^2 -4(81\times -49.23)} }{2(81)}\\\\R = 1.74 \ ohms \ \ or \ \ -0.35 \ ohms[/tex]
Thus, the two values of R when the power dissipated is 81 W are 1.74 ohms or -0.35 ohms.
Learn more here:https://brainly.com/question/15272925
