phy f.5 x1

Let V be the e.m.f. of the battery, R be the resistance of a bulb.
Equivalent resistance of the ABC
= [1/(R + R) + 1/R]-1
= 2R/3
Voltage drawn by D
= RV / (2R/3 + R)
= 3V/5
Voltage drawn by C = V - 3V/5 = 2V/5
Voltage drawn by A = voltage drawn by B = V/5
Power of D, V' = (3V/5)2/R = 9V2/25R
Power of C = (2V/5)2/R = 4V2/25R = 4V'/9
Power of A = Power of B = (V/5)2/R = V2/25 = V'/9
So, the brightness of D to that of C to that of A and B is 9 : 4 : 1. With D is the brightest, A and B be the dimmest.

b. When B burns out, the circuit through A and B breaks. So, A and B goes out.
C will increase in brightness, D will decrease in brightness, with they are at the same brightness.