>>12159569In a series circuit, the sum of the voltage drops are equal to the voltage source.
Each of those bulbs could not be 230 V. That would have to be the sum of the voltage drop and not for each bulb.
But putting aside that error, the bulbs are incandescent. Their light is produced by resistance heating, which is governed by the equation
P = R×I^2
and the current across a series circuit is constant.
P(L1) = R(L1)×I^2 and P(L2) = R(L2)×I^2
The actual voltage drop would be the sum of the resistance of the two bulbs times the current
V(drop) = [R(L1) + R(L2)] × I
Now there are three equations and three unknowns.
60 = R(L1) × I^2
100 = R(L2) × I^2
230 = [R(L1)+R(L2) × I
The current is 0.696 amps, the resistance of the L1 and L2 bulbs are 124 and 206.6 ohms, respectively. The voltage drop for L1 and L2 are 86.3 and 143.7 volts, respectively, and they add up to 230 volts
But these calculations are redundant since the 100 watt bulb will be the brighter of the two. I just did it for the hell of it.
