CHAPTER 8
Conceptual questions
Q14: All the current flows through the short circuit. No more current in the light bulbs. It is like a parallel circuit with three branches, on with a resistance of 0 ohms: the current chooses the path of least resistance.
Q15: In parallel, each light bulb has the same power as if it was alone in the circuit. The light bulb are brighter. If you do the calculations:
In series, P = (E/2)^2 / (1 ohms) = E^2/4 . In parallel, P = (E^2) / 1 ohm = E^2. The light bulbs are four times brighter in parallel.
Q16: B is only used as a conductor, as if it were a piece of wire. The current does not circulate in the filament. It is dark.
Q17: Bulb A needs to share the voltage provided by the battery with B. And because of B, the total resistance in the circuit is doubled, so the current delivered by the battery is divided by two. The electric power in A is therefore divided by 4 when B is added.
Q19: Power = (Delta V)^2 / R. Increasing the resistance in B does not change the voltage across A, nor the resistance of A. So nothing changes for A.
Q12: Calculate the equivalent resistance of the parallel association of resistances. The circuit to the left has two 1 ohm bulbs, each get 4.5V. The current is 9V/ 2 ohm = 4.5 A. P1 = P2 = 4.5 A x 4.5 V = 20.25W. In the circuit to the right: one resistance of 1 ohm (gets 6 V) and one resistance of 0.5 ohm (gets 3V). Current is 9V/1.5 ohm = 6A. p3 = 6V x 6A = 36 W and P4 = 3V x 6A= 18 W. Bulb #3 is the brightest.
Q14: All the current flows through the short circuit. No more current in the light bulbs. It is like a parallel circuit with three branches, on with a resistance of 0 ohms: the current chooses the path of least resistance.
Q15: In parallel, each light bulb has the same power as if it was alone in the circuit. The light bulb are brighter. If you do the calculations:
In series, P = (E/2)^2 / (1 ohms) = E^2/4 . In parallel, P = (E^2) / 1 ohm = E^2. The light bulbs are four times brighter in parallel.
Q16: B is only used as a conductor, as if it were a piece of wire. The current does not circulate in the filament. It is dark.
Q17: Bulb A needs to share the voltage provided by the battery with B. And because of B, the total resistance in the circuit is doubled, so the current delivered by the battery is divided by two. The electric power in A is therefore divided by 4 when B is added.
Q19: Power = (Delta V)^2 / R. Increasing the resistance in B does not change the voltage across A, nor the resistance of A. So nothing changes for A.
CHAPTER 9
