A battery of emf 24V and internal resistance 4Ω is connected to a resistor of 32Ω. What is the terminal p.d of the battery?

Assessment: WAEC SSCE - Physics - 1988 (Objective) Subject: Physics

Question 1 Report

A battery of emf 24V and internal resistance 4Ω is connected to a resistor of 32Ω. What is the terminal p.d of the battery?
Answer Details
The terminal p.d of a battery is the voltage difference between its positive and negative terminals when it is connected to a circuit. In this case, we have a battery of emf 24V and internal resistance 4Ω connected to a resistor of 32Ω. When a current flows through the circuit, there will be a voltage drop across the internal resistance of the battery, which will reduce the voltage available at the terminals. To calculate the terminal p.d of the battery, we can use Ohm's law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance. First, we need to calculate the current in the circuit using the total resistance, which is the sum of the internal resistance and the external resistance: Rtotal = Rinternal + Rexternal = 4Ω + 32Ω = 36Ω Using Ohm's law, we can calculate the current in the circuit: I = V/Rtotal I = 24V/36Ω I = 0.67A Now we can calculate the voltage drop across the internal resistance using Ohm's law: Vinternal = IRinternal Vinternal = 0.67A x 4Ω Vinternal = 2.68V Finally, we can calculate the terminal p.d of the battery by subtracting the voltage drop across the internal resistance from the emf of the battery: Vterminal = emf - Vinternal Vterminal = 24V - 2.68V Vterminal = 21.32V (approx.) Therefore, the terminal p.d of the battery is approximately 21.3V, which is the correct option.

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