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.
It depends on the type of interference.
For constructive interference, add the amplitudes to get |35 + 41| = 76 units.
For destructive, subtract them |35 - 41| = 6 units
Answer:
The work could be either positive or negative, depending on the direction the object moves
Explanation:
Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = 
C' = 
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
The energy of the capacitor, E is given by;
