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A resistor with resistance Ris connected to a battery that has emf 12.0 V and internal resistance r=0.40Ω. For what two values of R will the power dissipated in the resistor be 80.0 W ?

Short Answer

Expert verified

The two values of Rare 0.2Ωand 0.8Ω.

Step by step solution

01

Define the ohm’s law, resistance (R) and power (P).

Consider the formula for the Ohm’s law.

V=IR

Here, Iis current in ampere A,R is resistance in ohms role="math" localid="1655724796140" Ωand Vis the potential difference volt V.

Consider the power role="math" localid="1655724850722" (P)is the product of potential difference(V) and the current(I) and is given as follows:

P=I2R

Consider the current if the emf is ε, the internal resistance is r and the load resistance is Ris as follow:

I=εR+r

02

Determine the two resistances.

Consider the given, emf, internal resistance and the power consumed.

ε=12.0Vr=40ΩP=80.0W

Derive the equation to determine the resistance.

P=I2RP=εR+r2RP=ε2RR2+2Rr+r2R2=2Rr-ε2RP+r2=0

Substitute the values of P,randεin the equation.

R2+2R0.4-122R80+0.42=0R2-R+0.16=0R2-0.2R-0.8R+0.16=0R-0.2R-0.8=0

Solve further as,

R=0.2ΩR=0.8Ω

Hence, the two values of Rare 0.2Ωand 0.8Ωwill dissipated the power of 80.0 W .

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