# Circuit Question

1. Oct 30, 2005

### Erwin Schrodinger

A single resistor (R) is connected to a battery. The current is (I) and the power dissipated as heat is (P). The circuit is changed by doubling the emf (ε) of the battery while R is kept constant.

After the change, the current is how many times greater than the original?

original:
V = IR
V/R = I

after the change:
V = IR
2V = IR
2V/R = I
2I = I

The current is 2 times greater than the original.

After the change, the power dissipated in R is how many times greater than the original?

original:
P = IV

after the change:

P = IV
P = 2I(2V)
P = 4IV
P` = 4P

The power dissipated in R is 4 times greater than the original.

Are my answers and explainations correct? My textbook doesn't show the answers for even numbers.

Last edited: Oct 30, 2005
2. Oct 30, 2005

### cepheid

Staff Emeritus
Yes, you are right. You did the algebra, so I'm not sure what more confirmation you need. But think about it conceptually: in a resistor that obeys ohm's law, there is a linear relationship between voltage and current. Hence, if you double the voltage supplied to the resistor, you double the current drawn by it. In other words, this is a question you can do by inspection, too. Same with the power. You should know that the power dissipated in a resistor is given by:

P = V2/R

or

P = I2R

(Do you now how I got these formulae? Eliminate either V or I from P = IV by substituting one expressed in terms of the other.)

..and so this is not a linear, but a quadratic relationship. Doubling either V or I will therefore quadruple the power dissipated. Those formulae allow you to arrive at the answer more directly, but skips out an intermediate step.

However, the way you arrived at the answer is very instructive and useful too! You thought it through step by step: "If I double the voltage, I also end up doubling the current (as we just showed), and so the power quadruples, since it is the product of both."

You can see that there is no difference between the two methods really. Both make use of the relationship between current and voltage, one just does so a little less directly.