Applying Kirchoff Junction, rule what happens to the power source in current, when the parallel circuit has two branches, each with a resistor, R? I want to do senses answer the question. Explain your reasoning.

Answer 1

According to Kirchoff's Junction Rule, the sum of currents entering a junction must be equal to the sum of currents leaving the junction.

In the parallel circuit with two branches, there are two junctions - one at the beginning where the power source is connected, and one at the end where the branches meet.

At the starting junction, the current from the power source splits into two parallel branches with resistors R. Let's say the current from the power source is I, then according to the junction rule, the sum of currents entering the junction is I, and the sum of currents leaving the junction is also I, as there are no other paths for the current to go.

Now, let's consider one of the branches. The current entering this branch from the starting junction is I, and according to Ohm's Law (V = IR), the voltage across the resistor R in this branch is given by V1 = IR (where R is the resistance of each branch). The power consumed in this branch can be calculated as P1 = IV1 = I(IR) = I^2R.

Similarly, for the other branch, the current entering this branch from the starting junction is I, and the voltage across the resistor R in this branch is also V2 = IR (as both branches have the same resistance). The power consumed in this branch can be calculated as P2 = IV2 = I(IR) = I^2R.

Hence, the total power consumed in the parallel circuit with two branches is given by the sum of the power consumed in each branch, which can be written as: P_total = P1 + P2 = I^2R + I^2R = 2I^2R.

Therefore, the power consumed in the circuit is directly proportional to the square of the current (I^2), which means that if the current (I) increases, the power consumed by the parallel circuit will increase as well.