Power problem?

Question

A resistor R1 dissipates power P when connected to generator.If a resistor R2 is connected in series with R1 the power dissipated by R1-
1. increases
2. decreases
3. remains same
4. none of these

Answer 1

remains the same.

since P= (i^2)r,
and i (current) is constant in a series circuit,
then power dissipated by R1 is the same.

Answer 2

2. decrease
fairly simple to work out using ohms law and Watts law as follows:
R = resistance
E = voltage
I = current
P = power

Pie charts:
ohms law:
E
-------
I | R
watts law:
P
------
E | I
in both pie charts put your finger over what you want to find then either multiply or divide whichever is indicated:

Rt = R1 + R2 (total resistance of a series circuit will be all the individual resistors added together:
I = E/Rt so the more resistance the less current
Pt = E * I
power dissipated by each resistor:

P1 = E1 (voltage drop across R1) * I(current)
P2 = E2 (voltage drop across R3) * I (current)
Another way to express it:
E = I*R so
P = (I*R) * I
collecting the terms will give you:
P = I*I *R
multiplying I*I gives you I^2 (current squared)
therefore it
P = I^2 * R

What regreg was thinking of was a parallel circuit not a series circuit common mistake.

Answer 3

2 because the current will have decreased due to the extra resistance

Answer 4

2.decreases
coz d current decreases

Answer 5

2. decreases