The circuit shown in Figure P28.38 has been connected for a...

Key Concepts

Steady-State Behavior of Capacitors in DC Circuits

When a capacitor has been connected to a DC source for a long time, the circuit reaches a steady state in which the capacitor becomes fully charged and no current flows through it. In this state, the capacitor behaves like an open circuit. This concept is crucial for analyzing circuits with capacitors in DC conditions, as it allows for simplifications in circuit analysis by replacing the capacitor with an open branch when determining steady-state voltages and currents.

RC Time Constant

The RC time constant, denoted by ? (tau), is the product of the resistance (R) and the capacitance (C) in a circuit. It characterizes the rate at which a capacitor charges or discharges. The time constant is fundamental in predicting the transient response of RC circuits; for example, it helps determine the time required for a capacitor to discharge to a specific fraction of its initial voltage, as it defines the exponential rate of decay in these circuits.

Exponential Discharge

The discharge of a capacitor through a resistor follows an exponential decay process, where the voltage decreases by a fixed proportion over each time interval equal to one time constant. This principle allows for calculating the time needed for the capacitor's voltage to drop to a given fraction of its initial value. In analysis, the formula V(t) = V?e^(–t/RC) is used to assess how long the voltage takes to reach a particular level, making this concept integral to transient circuit analysis.

Transcript

00:02 Hi, the given circuit is having a combination of various resistors with a capacitor in the form of the arms of a rhombus.

00:17 The arms are in the form of resistors, four resistors and the capacitor joint along a diagonal of this rhombus.

00:32 Now the values of these resistors are this is 1 .00 om this is 8 .00 om this is 4 .0 om and finally here this is 2 .00 om then there is a battery providing a voltage of 10 volt to this circuit the value of capacitance is given as 1 .00 micro ferret now in the first part of the problem and before doing that we should name its vertices also so the naming of vertices of this circuit is a b c and d now in the first part of the problem after a long time, after making the circuit closed, the capacitor will become fully charged.

02:01 So there will be no current passing through it.

02:18 Hence we can ignore it as we have to find the potential difference across this capacitor in the first part of the problem.

02:35 So that potential difference actually will be equal to the difference between the the voltage at b and d so first of all we should find the voltages at point b and at points d to do this we reconstruct this circuit diagram so if we will ignore this capacitor now in that case 8 om and 2 om will be in series making it 10 om and this 1 om and 4 om will also be in series making it 5 om so if we redraw this circuit diagram we get it like this here this one 10 ome and in parallel with this here this is 5 om and both of them are in parallel with the battery providing a potential of 10 volt and as potential remains same in parallel so across both of these branches the potential will be same the voltage will be same so the current are i and i2 passing through these two branches may be known.

03:46 In this main circuit, this will be i1 and this will be i2...

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