Question

Show that the total effective resistance of a series combination of resistors is tje sum of individual resistance of the resistors in the combination​

Answer 1

❤❤Answer:

Answer:Let's assume that we have been given n number of resistors combined in series.

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:The basic thing to remember that Potential Difference is divided along the resistances in a series combination. But Current remains same along all the resistance.

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:The basic thing to remember that Potential Difference is divided along the resistances in a series combination. But Current remains same along all the resistance.V(total) = V1 + V2 + ....n \: timesV(total)=V1+V2+....ntimes

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:The basic thing to remember that Potential Difference is divided along the resistances in a series combination. But Current remains same along all the resistance.V(total) = V1 + V2 + ....n \: timesV(total)=V1+V2+....ntimes= > I(R \: eq.) = I(R1) + I(R2) + ....n \: times=>I(Req.)=I(R1)+I(R2)+....ntimes

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:The basic thing to remember that Potential Difference is divided along the resistances in a series combination. But Current remains same along all the resistance.V(total) = V1 + V2 + ....n \: timesV(total)=V1+V2+....ntimes= > I(R \: eq.) = I(R1) + I(R2) + ....n \: times=>I(Req.)=I(R1)+I(R2)+....ntimesCancelling I term :

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:The basic thing to remember that Potential Difference is divided along the resistances in a series combination. But Current remains same along all the resistance.V(total) = V1 + V2 + ....n \: timesV(total)=V1+V2+....ntimes= > I(R \: eq.) = I(R1) + I(R2) + ....n \: times=>I(Req.)=I(R1)+I(R2)+....ntimesCancelling I term := > R \: eq. = R1 + R2 + ...n \: times=>Req.=R1+R2+...ntimes

Answer:Let's assume that we have been given n number of resistors combined in series.We have to show that pts equivalent resistance is equal to sum of individual resistance.Concept:The basic thing to remember that Potential Difference is divided along the resistances in a series combination. But Current remains same along all the resistance.V(total) = V1 + V2 + ....n \: timesV(total)=V1+V2+....ntimes= > I(R \: eq.) = I(R1) + I(R2) + ....n \: times=>I(Req.)=I(R1)+I(R2)+....ntimesCancelling I term := > R \: eq. = R1 + R2 + ...n \: times=>Req.=R1+R2+...ntimesSo mathematically we can say that :

$\boxed{ \red{ \bold{ \huge{ R \: eq. = \sum \:( R )}}}} </em></p><p><em>[tex]\boxed{ \red{ \bold{ \huge{ R \: eq. = \sum \:( R )}}}}$