Cracking the SAT Physics Subject Test

For the 4 Ω resistor: P = IV = (2 A)(8 V) = 16 W

For the 3 Ω resistor: P = IV = ( A)(4V) = W

For the 6 Ω resistor: P = IV = ( A)(4V) = W

So the resistors are dissipating a total of

If the resistors are dissipating a total of 24 J every second, then they must be
provided with that much power. This is easy to check: P = IV = (2 A)(12 V) = 24
W.

Questions 4-7

For the circuit below,