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.(adsbygoogle = window.adsbygoogle || []).push({});

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

original:

V = IR

V/R = I

after the change:

V` = I`R

2V = I`R

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` = I`V`

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.

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# Homework Help: Circuit Question

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