In the circuit of Fig. 26.44, each resistor represents a light bulb. Let R1= R2=R3=R4=4.50 Ωand

In the circuit of Fig. 26.44, each resistor represents a...

Key Concepts

Power Dissipation in Resistors

The concept of power dissipation in resistors is crucial for understanding how electrical energy is converted into heat or light in a circuit. The power dissipated by a resistor can be calculated using formulas such as P = I²R or P = V²/R. This helps in determining how brightly a component like a light bulb might glow, based on the amount of power it consumes.

Kirchhoff’s Circuit Laws

Kirchhoff’s circuit laws, including the Current Law (KCL) and the Voltage Law (KVL), are essential for analyzing complex circuits. KCL states that the total current entering a junction equals the total current leaving it, while KVL implies that the sum of voltage changes around any closed loop must equal zero. These rules help in setting up equations to solve for unknown currents and voltages in the circuit.

Circuit Modification and Its Effects

When elements are added or removed from a circuit, the overall behavior changes significantly. Understanding circuit modifications involves analyzing how the total resistance, current distribution, and voltage drops are affected by changes in the circuit topology. This is important for predicting the performance of the circuit under different configurations, such as how the removal of a resistor influences the brightness of the remaining bulbs.

Ohm's Law

Ohm's Law is the foundational principle in circuit analysis that relates voltage (V), current (I), and resistance (R) through the equation V = IR. This law is essential for determining how voltage is divided among different components and how current flows through them, and it forms the basis for analyzing the electrical behavior of circuits.

Series and Parallel Resistances

Understanding series and parallel configurations is critical for analyzing circuits. In series circuits, resistances add up, affecting the total resistance and thus the current flow. In parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances, affecting the distribution of current and the overall voltage drop across the network.

Transcript

00:01 In the given problem, there is a circuit diagram having few bulbs arranged in mixed grouping of series and parallel.

00:16 Like shown here, this is the resistance r1, the bulb having a resistance of r1.

00:23 Then this is another bulb having a resistance r4 then here this is r2 and finally another bulb is joined diagonally here having a resistance r3 and a common source of emf is applied here, positive and a negative potentials as shown here.

01:03 Each resistor is representing a light bulb and all the resistors are having the same value as per the given question.

01:14 R1 is equal to r2 is equal to r3 is equal to r4 and all are having a value of 4 .50 and the emf applied is having a value of 9 .00 volt in the first part of the problem we have to find the current in each bulk for which first of all we need to find the net resistance of this combination for which this r2 and r3 and r4 are in parallel r2 r3 and r4 are in parallel r2 r3 and r4 are in a parallel combination.

02:04 So their net resistance if rp, that rp is found using the expression for parallel combination as 1 upon rp is equal to 1 upon r2, plus 1 upon r3 plus 1 upon r4.

02:20 And it becomes 1 by 4 .5 plus 1 by 4 .5 plus 1 by 4 .5 is equal to.

02:29 2 3 5 4 .5.

02:32 So this rp here comes out to be 1 .5 om.

02:40 Now this rp is in series with r1.

02:53 So net resistance, net effective resistance of this combination is r1 plus rp.

03:01 Means 4 .50 oom plus 1 .50 om which comes out to be 6 .00 om so current the net current passing through this circuit i which will also be equal to i1 means the current passing through this resistance r1 as current remains same in series so i or we can say the current passing through resistance r1 is given by home slow as e by r effective the net resistance here this e is 9 volt 9 .00 volt divided by this 6 .00 home so this current here comes out to be 1 .5 ampere this is one of the answer the current passing through resistance now we have to find current in each and every resistance so if we give this square as a name a and p these terminals are named as ab so the current in this parallel combination will also be same ip or iab will also be same as this combination is in series with r1 and the current will be 1 .5 am here.

04:39 Now this current 1 .5 ampere will be divided into three parts as i 2, i 3 and i 4 and as all the three resistors are equal to the current will be divided into equal parts so i 2 will be equal to i 3 will be equal to i 4 which is 1 third of the total current so we can say 0 .5 ampere so the three currents are having a value of 0 .5 ampere each and the net current or the current through first bulb is 1 .5 amp now in the next part of the problem in part b we have to find the power dissipated in each bulb so to find the power dissipated in each bulb so to find the dissipated we will use the expression for power which is i square into r so for p1 this is i1 square into r1 which becomes 1 .5 to the whole square multiplied by 4 .5 in watt which will come out to be 10 .1 to 5 watt then for p2 and p3 and p4 means all the three bulbs arranged in parallel the power dissipated will be same given by 0 .5 square into 4 .5 means each will be having a value of 1 .1 to 5 watt so here this will be the brightest so this is the answer for the second part of the problem part b now in third part the bulb r4 is now removed from the search leaving a brake in the wire at its position so if we remove this r4 from the circuit let it be removed so now only two resistors are in parallel here this is r2 and r3 so now removing r4 the net resistance of parallel combination will come out to be 4 .5 divided by 2 om or we can say this is 2 .25 om.

07:28 So now net resistance, effective resistance of the combination will be just 2 .25 plus r1 which is 4 .5 which will come out to be 6 .75 om.

07:41 So now if we have to find the net current passing through this combination it will come out to be i or it will be the same current which is passing through r1 which is i equals to i1 divided by emf by r so now are effective so now this is again 9 .0 volt divided by 6 .75 so this current comes out to be 1 .3 amp here and similarly this current further will be divided into two equal parts across r2 and r3 so this will be i2 and i3 which will be 1 .33 divided by 2 and it comes out to be 0 .67 ampe.

08:32 So these are the two answers for third part of the problem.

08:37 Now in the fourth part of the problem we have to find the power dissipated in these bulbs now.

08:48 So as far as p1 is concerned the power consumed...

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