Answer
(a) decrease
(b) increase
Work Step by Step
(a) Since the resistors are connected in series, the resultant resistance is 3R. When the switch is closed, the light is shorted out. It shows that the intensity of light 2 decreases.
(b) We know that when the switch is closed, light 2 is shorted out and thus the equivalent resistance is decreased from 3R to 2R. As $P=\frac{\mathcal{E}^2}{R}$, this equation shows that the power dissipation is inversely proportional to the resistance. We conclude that the intensity in light 1 and 3 increases due to the lower resistance.
