A battery of e.m.f 10V an internal resistance 2Ω is connected to a resistance of 6Ω. Calculate the p.d

Answer Details

The potential difference (p.d) across a resistor in a circuit can be calculated using Ohm's law which states that the p.d is equal to the product of the current flowing through the resistor and the resistance of the resistor. In this circuit, the battery has an e.m.f of 10V and an internal resistance of 2Ω. Therefore, the p.d across the battery terminals is 10V. Since the resistor of 6Ω is connected in the same series with the battery, the current flowing in the circuit will cause a voltage drop across the internal resistance of the battery. This means that the p.d across the resistor will be less than the 10V. The current flowing in the circuit can be calculated using the formula I = E/(R + r), where E is the e.m.f of the battery, R is the resistance of the external resistor, and r is the internal resistance of the battery. Substituting the given values, we get: I = 10V/(6Ω + 2Ω) = 1.25A The p.d across the resistor can then be calculated using Ohm's law: p.d = IR = 1.25A x 6Ω = 7.5V Therefore, the answer is 7.50V.