Question

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 ?

Answer 1

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.)