Question

A 16.05 - V battery with internal resistance \(R_{j}\) is being charged by a battery charger that is capable of delivering a current \(i=6.041 \mathrm{~A}\). The battery charger supplies an emf of \(16.93 \mathrm{~V}\). What is the internal resistance, \(R_{\mathrm{j}}\), of the battery?

Short answer

Answer: The internal resistance of the battery is approximately 0.1456 Ω.

Step-by-step solution

Understanding Kirchhoff's Voltage Law for a simple circuit

In this situation, we have a simple circuit consisting of a battery charger and a battery with an internal resistance. According to Kirchhoff's Voltage Law (KVL), the sum of the potential differences (voltages) around any closed loop in a circuit is equal to zero. In this case, the loop consists of the battery charger, the battery, and its internal resistance.

Apply Kirchhoff's Voltage Law to the circuit

Applying KVL to the circuit, we have: \(V_{charger} - V_{battery} - i \cdot R_{j} = 0\) Where \(V_{charger}\) is the emf of the battery charger (\(16.93V\)), \(V_{battery}\) is the emf of the battery (\(16.05V\)), \(i\) is the current flowing in the circuit (\(6.041A\)), and \(R_{j}\) is the internal resistance of the battery that we need to find.

Rearrange the equation to isolate \(R_{j}\)

We need to isolate \(R_{j}\) on one side of the equation to solve for it. Rearranging the equation gives us: \(i \cdot R_{j} = V_{charger} - V_{battery}\) Now, we can divide both sides by the current \(i\) to obtain the internal resistance of the battery: \(R_{j} = \frac{V_{charger} - V_{battery}}{i}\)

Substitute the given values and solve for \(R_{j}\)

Now, we can plug in the given values for \(V_{charger}\), \(V_{battery}\), and \(i\): \(R_{j} = \frac{16.93V - 16.05V}{6.041A}\) \(R_{j} = \frac{0.88V}{6.041A}\) Calculating the internal resistance, we get: \(R_{j} \approx 0.1456 \Omega\) Therefore, the internal resistance of the battery is approximately \(0.1456 \Omega\).