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Chapter 4: Q61P (page 845)

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes10.20 V . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short Answer

Expert verified
  1. The internal resistance of the battery is0.36Ω.
  2. The emf of the battery is 8.94 V .

Step by step solution

01

Define the ohm’s Law.

Consider the real source of emf has internal energy then its terminal potential differenceVab is as follows:

Vab=ε-IR ….. (1)

02

Determine the internal resistance.

(a)

Consider the given expression for the current and the voltage.

Vba=8.40VIba=1.5A

Substitute the values in the equation (1).

8.40=ε-(1.5)(R)ε=8.40+1.5r

Consider the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V.

Substitute the values in the equation V=ε+lrand solve.

10.20=(8.30+1.5r)+3.5r5r=1.8r=0.36Ω

Hence, the internal resistance of the battery is0.36Ω

03

Determine the emf of battery.

(b)

Given: the value of the internal resistance is 0.36Ω.

Substitute 0.36Ωfor r in the equation ε=8.40+1.5r.

ε=8.40+1.5×0.36=8.94V

Hence, the emf of the battery is 8.94 V.

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