Answer the following questions 1. A simple circuit is made up of a 20 V battery, conducting wires and a resistor. If a current of 5 A flows through the wires, resistor and battery: 1.1 Calculate the amount of electrical potential energy that the battery supplies to 3 C of charge. 1.2 How much of charge flows through the resistor in 15 seconds? .3 How many electrons move through the resistor in 15 seconds? .4 How much charge must be given out by the battery to supply 200 J of energy? 5 How long will it take the battery to transfer 200 J of energy?

Let's solve these questions step by step: ### 1.1 Calculate the amount of electrical potential energy that the battery supplies to 3 C of charge. The electrical potential energy (E) can be calculated using the formula: \[ E = V \times Q \] where \( V \) is the voltage (20 V) and \( Q \) is the charge (3 C). So, \[ E = 20 V \times 3 C = 60 J \] The battery supplies 60 J of electrical potential energy to 3 C of charge. ### 1.2 How much charge flows through the resistor in 15 seconds? The current (I) can be used to find charge (Q) using the formula: \[ Q = I \times t \] where \( I = 5 A \) and \( t = 15 s \). So, \[ Q = 5 A \times 15 s = 75 C \] A total of 75 C flows through the resistor in 15 seconds. ### 1.3 How many electrons move through the resistor in 15 seconds? To find the number of electrons, we can use the charge of a single electron, which is approximately \( 1.6 \times 10^{-19} C \). The number of electrons (n) can be calculated using: \[ n = \frac{Q}{e} \] So, \[ n = \frac{75 C}{1.6 \times 10^{-19} C} \approx 4.69 \times 10^{20} \text{ electrons} \] Approximately \( 4.69 \times 10^{20} \) electrons move through the resistor in 15 seconds. ### 1.4 How much charge must be given out by the battery to supply 200 J of energy? Using the formula for electrical potential energy: \[ E = V \times Q \], we rearrange it to find \( Q \): \[ Q = \frac{E}{V} \] So, \[ Q = \frac{200 J}{20 V} = 10 C \] The battery must give out 10 C of charge to supply 200 J of energy. ### 1.5 How long will it take the battery to transfer 200 J of energy? Using the relationship between charge and current: \[ Q = I \times t \], we can find time (t): First, we know \( Q = 10 C \) (from question 1.4) and \( I = 5 A \): \[ t = \frac{Q}{I} = \frac{10 C}{5 A} = 2 s \] It will take the battery 2 seconds to transfer 200 J of energy.