89.4k views
1 vote
A resistor with resistance R is connected to a battery that has emf 13.0 V and internal resistance r = 0.350 Ω.

For what two values of R will the power dissipated in the resistor be 77.0 W ?

User Thilak Rao
by
8.3k points

1 Answer

3 votes

Answer:

Approximately
1.41\; {\rm \Omega} and approximately
0.0870\; {\rm \Omega}.

Step-by-step explanation:

Since the emf of
\epsilon = 13.0\; {\rm V} is divided between
r = 0.350\; {\rm \Omega} and
R, the current in this circuit would be:


\begin{aligned}I &= (\epsilon)/(R + r)\end{aligned}.

The resistor
R would dissipate:


\begin{aligned}P &= I^(2)\, R \\ &= \left((\epsilon)/(R + r)\right)^(2)\, R \\ &= (\epsilon^(2)\, R)/((R + r)^(2))\end{aligned}.

Rearrange this equation to obtain:


P\, (R + r)^(2) = \epsilon^(2)\, R.


P\, (R^(2) + 2\, r\, R + r^(2)) = \epsilon^(2)\, R.


(P)\, R^(2) + (2\, P\, r - \epsilon)\, R + (P\, r^(2)) = 0.

Given that
P = 77.0\; {\rm W}:


(77)\, R^(2) + (-116.1)\, R + (9.4325) = 0.

Solve this quadratic equation of
R using the quadratic formula to obtain:


R \approx 1.41\; {\rm \Omega} or
R \approx 0.0870\; {\Omega}.

(Substitute each value into the equation
P = (\epsilon^(2)\, R) / ((r + R)^(2)) to verify the results.)

User Moein Kameli
by
7.7k points