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An idealized voltmeter is connected across the terminals of a15.0-Vbattery, and arole="math" localid="1655719696009" 75.0-Ω appliance is also connected across its terminals. If the voltmeter reads11.9V (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

Short Answer

Expert verified

(a) The power dissipated by the appliance is1.89W.

(b) The internal resistance of the battery is19.62Ω.

Step by step solution

01

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

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

V=IR

Where,Iis current in ampere A, is resistance in ohms Ωand Vis the potential difference volt V.

If real source of emf εhas internal energy rthen its terminal potential difference Vabdepends upon current.

So, formula for the terminal potential difference is:

Vab=ε-Ir

The ratio of V toI for a particular conductor is called its resistanceR

R=VIorpLA

Where, pis resistivityΩm,Lis length in mand Ais area in m2.

The power (P)is the product of potential difference (V)and the current(I) .

P=VIorI2RorV2R

02

Determine the power.

Given that,

V=11.9VR=75Ω

The power dissipated in the circuit is:

P=V2R=11.9275=1.89W

Hence, the power dissipated by the appliance is

1.89W

03

Determine the internal resistance.

Given that,ε=15.0V

The internal resistance of the battery is:

r=ε-VIr=ε-VVR

Substitute the values of variables

r=15-11.911.975=3.10.158=19.62Ω

Hence, the internal resistance of the battery is19.62Ω.

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