WAEC SSCE - Physics - 2000 (Objective)

Question 1 Report

In the diagram given, the current passing through the 6\(\Omega\) resistor is 1.5A. Calculate the terminal p.d. of the battery in the diagram

7.50V

9.00V

10.80V

1.25V

Answer Details

To solve this problem, we need to use Kirchhoff's voltage law (KVL) which states that the sum of the potential differences around any closed loop in a circuit is zero. Starting from the positive terminal of the battery and moving clockwise around the circuit, we encounter a potential drop across the 6Ω resistor and a potential rise across the battery. Using Ohm's law, we can calculate the potential drop across the 6Ω resistor as: V = IR = 1.5A × 6Ω = 9V This means that the potential difference across the battery is: V_battery = V_6Ω + V_rise = 9V + V_rise We don't know the potential rise yet, but we can use KVL to find it. Moving clockwise around the circuit, we encounter a potential rise across the battery, a potential drop across the 4Ω resistor, and another potential drop across the 2Ω resistor. Using Ohm's law and KVL, we can write: V_rise - I × R_4Ω - I × R_2Ω = 0 where I is the current passing through the 6Ω resistor, which is also the current passing through the entire circuit. Substituting the given values, we get: V_rise - 1.5A × 4Ω - 1.5A × 2Ω = 0 V_rise = 1.5A × 6Ω = 9V Therefore, the potential difference across the battery is: V_battery = V_6Ω + V_rise = 9V + 9V = 18V So the terminal potential difference of the battery in the circuit is 18V. Therefore, the correct option is 10.80V.

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