For the circuit shown in Figure P 23.72, find the current through and the potential difference a

For the circuit shown in Figure P 23.72, find the current...

Key Concepts

Circuit Analysis Techniques

Circuit analysis techniques involve systematic methods such as mesh analysis and node voltage analysis. These approaches leverage fundamental laws like Ohm's Law and Kirchhoff’s Laws to break down complex circuits into simpler parts, enabling the calculation of unknown currents and voltages across various elements.

Equivalent Resistance

Equivalent resistance is the single resistance value that can replace a group of resistors in a circuit without changing the overall current or voltage characteristics. This concept is central to reducing complex circuits into simpler forms for easier analysis using series and parallel resistor rules.

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits stating that the voltage across a resistor is directly proportional to the current flowing through it, with the constant of proportionality being the resistance. This relationship, expressed as V = IR, is crucial for calculating current or voltage in resistive elements.

Kirchhoff’s Circuit Laws

Kirchhoff's Circuit Laws consist of two key principles: Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL). KCL mandates that the total current entering a junction equals the total current leaving, while KVL states that the sum of all electrical potential differences around a closed loop must be zero. These laws are essential for analyzing current and voltage distribution in complex circuits.

Series and Parallel Circuits

Understanding series and parallel arrangements of resistors is vital in circuit analysis. In series circuits, resistors are connected end-to-end, and their total resistance is the sum of individual resistances. In parallel circuits, resistors share common nodes, and the overall resistance can be found using the reciprocal formula. This concept helps simplify and solve networks of resistors.

Transcript

00:02 The given combination in this problem is having so many resistors joined in series and in parallel.

00:15 In the mixed grouping, first of all, this is 4 -on and here this is 24 -oom.

00:25 Then this is 6 -oam, which is joined with the two resistors in parallel.

00:36 8 oom and 24 om again.

00:44 Then finally all these three terminals are joined with the negative terminal of the battery which is having an emf of 24 volt.

00:59 So in the problem we have to find the current passing through each resistor and the potential drop taking place across each resistor so first of all, we will find the equivalent resistance of this combination.

01:15 To do this, first of all, these two resistors are in parallel combination.

01:21 So their value will be given as 8 into 24 in the numerator their product and in denominator their sum which comes out to be 8 into 24 divided by 32 which is 3.

01:50 It comes out to be 6 oom.

01:57 Now this 6 om is in series combination with another 6 ome.

02:04 So their equivalent resistance will come out to be now 6 plus 6 means 12 oom.

02:13 Finally this 12 om is in parallel again with this 24 om giving the net resistance of 12 into 24 divided by 12 plus 24 and now it comes out to be 12 into 24 divided by 36 so this is 8 om and at last this 8 om is in series with this 4 om so equivalent resistance of this combination is 4 om plus 8 om means 12 om.

02:59 Hence, the net current coming out of the battery i is given as using oms law, emf divided by equivalent resistance.

03:10 So this is 24 by 12, which becomes 2 .0 ampere.

03:18 Now, as the same current will be passing through 4 oom, so current through 4 oom resistor is, is 2 .0 oom this is one of the answer and potential drop taking place across it using omslaw i into r so this is i 2 .0 oam ampere multiplied by resistance having a value of 4 oom so here it comes out to be 8 volt so this is complete set of answer for the first resistor now, if we consider this as a terminal a and this point as a terminal b, then as the potential 8 volt is getting dropped across 4 -on and the 24 -volt was applied.

04:24 So the remaining will be dropped across this combination ab...

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