1 consider the circuit in the figure below with v 600 v r1...

1. Consider the circuit in the figure below, with V= 6.00 V, R1= 7.00 Ohm, R2= 4.00 Ohm R3= 2.00 Ohm, R4= 5.00 Ohm and R5= 6.00 Ohm. a) What are the currents through R1, R2, R3, R4 and R5. b) What is the equivalent resistance of this combination of resistors? c) What is the potential difference between points b and c. d) What is the total power dissipated by these resistors?

Transcript

00:01 Hello, in this question we are given the circuit is like this.

00:04 All right.

00:05 So first of all, let us calculate the equivalent resistance for this circuit.

00:09 Now, these two r2 and r3 are connected in series.

00:12 So r2 and r3 can be replaced by one resistance and resistance of that resistor would be equal to six oms.

00:21 All right.

00:22 And now this six -home resistance and this r4, they are in parallel.

00:27 So they both can be now replaced by.

00:30 Even single register so register are equivalent would be equal to in this case six into five over six plus five so that would be equal to 2 .73 oms all right so now our circuit will be simplified to this much all right this is our 7 home this is our 6 home and this is our 2 .7 3 oms and there is a battery connected that is of 6 volt like this and now the equivalent register of the circuit would be what 7 plus 6 plus 2 .7 3 so it would be 15 .7 3 would be the equivalent resistance and the current would be what v is equal to i into r so i would be what v by r so 6 divided by 15 .73 would be our answers so i would be 0 .3814 all right mps here you have not uploaded the image of the circuit so i get it from the web there may be some error in the placement of resistance but accordingly you must have to solve the circuit all right so i have told you the method to solve the circuit values can vary like but method would be same now further what we need to do we need to calculate current in each of the register all right so this this now like this like this all right so from here how much current will be flowing 0 .3814 amp this would be flowing through r1 and from through r5 all right but in case of r2 r3 and r4 currents would be different and current would be in what in the ratio of in ratio such that the potential drop between this point and this point is same through both circuits from here and from here also.

02:42 So resistance here was what was resistance 5 om and resistance for this circuit was 6 om so correspondingly current would be vary this if this is i1 this is i 2 then i 1 into 5 is equal to i 2 into 6 and i 1 plus i 2 is equal to 0 .3814 so from here we can see that i2 would be equal to if i replace i1 by 1 .2 i2 all right so this would be 0 .3814 by 2 .2 so this i2 is coming out to be 0 .1734 and minus 0 .3814 this i2 will be coming out to be sorry i1 will be coming out to be 0 .2080.

03:39 So these are the currents in all the registers...

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