The total resistance in a series connection is equal to product of the individual resistance?

Answer 1

Yes. Power(P)=Current(I)xVoltage(E) or P=I x E, and since in a series circuit current is constant and voltage is additive then:

P(series)=Pr1+Pr2+Pr3........ Actually, power dissipated in series circuits is P = I^2 * R and in parallel power dissipated = V^2/R

Answer 2

The total resistance in a series connection is equal to SUMof the individual resistanceSay you want to measure the voltage across the series of these two resistors:

--- R1 --- R2 ---

Current I is the same in the two resistors. If you write 2nd kirchhoff's law, you get that the total voltage V is equal to VR1 + VR2.

For each resistor "k" you have VRk= Rk IRk.

By substitution, you have V = VR1 + VR2 = R1 I + R2 I = (R1 + R2) I = Req I

So, in the end, Req = R1 + R2

Answer 3

For resistors in a series (connected end to end), the resistances simply add together.

If the resistors are in parallel (side by side), things are a bit more complex: take the inverse of each resistance ( 1 / resistance ), and add them together; then take the inverse of the result. This looks like: 1 / ( 1/R1 + 1/R2 + ... + 1/Rn ).

If the arrangement is more complex than simple series or parallel resistances, then working out the resistances involves setting up and solving a system of linear equations.

Answer 4

What is the current running through the entire circuit? 3 amps

In a series circuit the current is the same in each resistor. The voltage across each resistor will be the current multiplied by the resistance.

Answer 5

The setup isn't quite clear. I'll assume you have two or more resistors in series, and in parallel. In parallel, each resistor gets the full voltage. For example, if the power source provides 10 V, each resistor gets 10 V. In series, each resistor gets only part of the voltage. For example, in the case of two identical resistors in series, each will get 5 V. As a result, less power will be dissipated. Exactly how much less will depend on how many resistors are used, and on the specific resistance values; you can calculate the power dissipated in specific cases with the formula P = V2/R, or P = I2R. Check the Wikipedia article on "Voltage divider", on how to calculate the voltage for individual resistors.

Answer 6

Yes, Voltage in series is additive. Formula Eapp=Er1+Er2+Er3..... -Kirchoffs Voltage Law: the sum of all voltage drops must equal the source voltage