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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` = 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.

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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