If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.
If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.
Answer:
Ratio of series current to parallel
= 1 : 8
Explanation:
Total resistance Rt
For series, Rt = 2+2+2+2 = 4ohms
For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2
1/Rt = 4/2, Rt = 2/4 ohms.
If we use a 1V battery, then,
I = V/Rt
I = 1/4 = 0.25 ampere for series arrangement.
I = 1/0.5 = 2 ohms.
Ratio of current of series to parallel = 0.25 : 2
= 1 : 8
The stage where atoms are spread out and bouncy is the gas stage.
Power is the amount of work done over a period of time. If you will put that into an equation, the formula of power will be:
P = W/t
Where:
P = power
W=work
t = time
Your problem already provides you with work and time so all you need to do is divide:
P = W/t
P = 50J/30s
P = 1.67 W