Cracking the SAT Physics Subject Test

From diagram , which has just one resistor, we can figure out the current.

Now we can work our way back to the original circuit (diagram ). In going from

to , we are going back to a series combination. Resistors in series share the

same current, so we take the current, I = 2 A, back to diagram. The current

through each resistor in diagram is 2 A.

Since we know the current through each resistor, we can figure out the voltage drop
across each resistor using the equation V = IR. The voltage drop across the 4 Ω
resistor is (2 A)(4 Ω) = 8 V, and the voltage drop across the 2 Ω resistor is (2 A)(2
Ω) = 4 V. Notice that the total voltage drop across the two resistors is 8 V + 4 V =
12 V, which matches the emf of the battery.

Now for the last step: going from diagram back to diagram. Nothing needs to

be done with the 4 Ω resistor; nothing about it changes in going from diagram to

, but the 2 Ω resistor in diagram goes back to the parallel combination.