Question

Question: The potential difference across the terminals of a battery is 840 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 becomes 10.20 V . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short answer

Answer

1. The internal resistance of the battery is$0.36\Omega$ .
2. The emf of the battery is 8.94 V .

Step-by-step solution

Define the ohm’s Law.

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

${\mathbf{V}}_{\mathbf{ab}}\mathbf{=}\mathbf{\epsilon }\mathbf{-}\mathbf{IR}$ ….. (1)

Determine the internal resistance.

(a)

Consider the given expression for the current and the voltage.

$$\begin{gathered} V_{ba}=8.40\text{V} \\ {I}_{ba}=1.5\text{A} \end{gathered}$$

Substitute the values in the equation (1).

$$\begin{gathered} 8.40=\epsilon -\left(1.5\right)\left(r\right) \\ \epsilon =8.40+1.5r \end{gathered}$$

Consider the current is in the reverse direction, the potential difference becomes .

Substitute the values in the equation $V=\epsilon +Ir$and solve.

$$\begin{gathered} 10.20=\left(8.30+1.5r\right)+3.5r \\ 5r=1.8 \\ r=0.36\Omega \end{gathered}$$

Hence, the internal resistance of the battery is 0.36 .

Determine the emf of battery.

(b)

Given: the value of the internal resistance is $0.36\Omega$.

Substitute for in the equation$\epsilon =8.40+1.5r$ .

$$\begin{gathered} \epsilon =8.40+1.5\times 0.36 \\ =8.94\text{V} \end{gathered}$$

Hence, the emf of the battery is 8.94 V.