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 81.0 W ?

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)

-----(1)

The power dissipated in the circuit is calculated as follows;

P = I²(R + r)

solve (1) and (2) together;

Thus, the two values of R when the power dissipated is 81 W are 1.74 ohms or -0.35 ohms.

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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

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

the area between the graph and the t axis