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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Most popular questions from this chapter

(See Discussion Question Q25.14.) An ideal ammeter A is placed in a circuit with a battery and a light bulb as shown in Fig. Q25.15a, and the ammeter reading is noted. The circuit is then reconnected as in Fig. Q25.15b, so that the positions of the ammeter and light bulb are reversed. (a) How does the ammeter reading in the situation shown in Fig. Q25.15a compare to the reading in the situation shown in Fig. Q25.15b? Explain your reasoning. (b) In which situation does the light bulb glow more brightly? Explain your reasoning.

In Figure below, if coil2 is turned90° so that its axis is vertical,

does the mutual inductance increase or decrease? Explain.

(See Discussion Question Q25.14.) Will a light bulb glow more brightly when it is connected to a battery as shown in Fig. Q25.16a, in which an ideal ammeter is placed in the circuit, or when it is connected as shown in Fig. 25.16b, in which an ideal voltmeter V is placed in the circuit? Explain your reasoning.

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage Vbaof the 16.0-V battery; (c) the potential difference Vacof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of , the induced emf in the second coil has magnitude 1.65×10-3V. (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the flux through each turn when the current in the first coil equals 1.20A? (c) If the current in the second coil increases at a rate of 0.360A/s, what is the magnitude of the induced emf in the first coil?

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