Current Electricity

The net resultant circuit will be solved as follows:

There is no change in current over time for a resistor. The best corresponding graph is .After a long time (steady state), the capacitor acts like an open circuit, meaning no current flows through it. The best corresponding graph is .After a long time, an inductor in a DC circuit behaves like a short circuit, meaning it allows the current to pass through with almost no resistance. The best corresponding graph is .

To understand the power dissipation in a cylindrical cable when it is cut in half, we begin by considering the formula for resistance: where: is the resistance, is the resistivity of the material, is the length, is the cross-sectional area.The power dissipated in a resistor due to a current is given by: Initially, the power dissipated in the full-length cable is . When the cable is cut into two equal halves, each half has a length of . Let's denote the power dissipated in one of these halves as .Since the resistance of each half becomes: and the current remains the same through both sections, the power dissipated in one half is: Thus, comparing and : Therefore, .This concludes that the power dissipated in each half of the cable is half of the original power, .

In a symmetrical circuit, equipotential points have the same voltage. Joining these points or removing resistors connecting these points does not change the voltage distribution or the overall behaviour of the circuit because no potential difference exists between them.They are symmetrically opposite each other across the central axis will have the same potential and can be joined to reduce the complexity of the analysis.The symmetry across the direct line from to shows that points , and are at the same potential. So, we can ignore the two resistances and the circuit becomes as shown above ( fig), this can be solved by adding in series and then in parallel. Otherwise, if all points are connected ( fig), add in parallel and then in series.Which will give us$ $