a) The indication "12 V" for each lamp means that when the lamp is connected to a voltage source, it will consume a voltage of 12 volts. The indication "0.6 A" for each lamp means that when the lamp is consuming 12 volts, it will draw a current of 0.6 amperes.
b) Since the fuse has negligible resistance, the voltage drop across it will be negligible. Therefore, the voltage across the lamps (L1 and L2) will be equal to the voltage of the generator (U), which is 24 V. Thus, U1 = U2 = 24 V.
c) Since the voltage across each lamp is within its specified range (12 V), and the current through each lamp is also within its specified range (0.6 A), we can conclude that the lamps are functioning normally.
d) The total current in the circuit is the sum of the currents through the lamps and the fuse. Since the lamps are identical and the voltage across them is the same, their currents will also be the same. Therefore, the total current I is given by:
I = I1 + I2 + Ifuse
Since each lamp consumes 12 V and draws 0.6 A, their resistance can be calculated as:
R = V/I = 12/0.6 = 20 Ω
Therefore, the total resistance of the circuit is:
Rtotal = 2R + Rfuse
Since the lamps are identical, their resistance adds up. We add the resistance of the fuse, which is negligible, so we can consider it as zero. Thus,
Rtotal = 2(20 Ω) + 0 Ω = 40 Ω
Now we can calculate the total current I using Ohm's law:
I = V/Rtotal = 24/40 = 0.6 A
Therefore, the current through the circuit is 0.6 A.