Question:

Consider the circuit shown below:

The voltage source, V = 12 V, is connected to a resistor of resistance R = 5 Ω. Determine the following:

a) The current flowing through the circuit.

b) The power dissipated by the resistor.

c) The energy dissipated by the resistor over a time interval of 5 seconds.

Answer:

a) To find the current flowing through the circuit, we can use Ohm's Law, which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor, divided by the resistance (R) of the resistor. Mathematically, this can be represented as:

I = V / R

Substituting the given values into the equation, we have:

I = 12 V / 5 Ω
I ≈ 2.4 A

Therefore, the current flowing through the circuit is approximately 2.4 A.

b) The power dissipated by a resistor can be calculated using the formula:

P = I^2 * R

Substituting the known values, we have:

P = (2.4 A)^2 * 5 Ω
P ≈ 28.8 W

Therefore, the power dissipated by the resistor is approximately 28.8 W.

c) The energy dissipated by a resistor is given by the formula:

E = P * t

where E represents energy, P represents power, and t represents time. Substituting the known values, we have:

E = 28.8 W * 5 s
E = 144 J

Therefore, the energy dissipated by the resistor over a time interval of 5 seconds is 144 J.