In a four-branch parallel circuit, there are 10mA of current in each branch. If one of the branches opens, the current in each of the other three branches is
a) 13.3 mA
b) 10 mA
c) 0A
d) either (b) or (c)
The book gives the answer as (b). I cannot figure out how (b) is the correct answer. According to my calculations the answers is (a). Please help me clarify this problem.
Calculation:
Total current is I total = (10 mA * 4) 40mA.
If one of the branches opens then the current is evenly split among 3 paths. The current for each path is Ix = (40 mA / 3) 13.3 mA. The problem does not specify the resistance for each path so I am assume each path has equal resistance.
Best Answer
You are assuming a constant current source and no load on each branch. In that case yes your answer would be right.
However, consider this circuit:
simulate this circuit – Schematic created using CircuitLab
The current through any one resistor (or branch of the circuit) will be \$I=V/R = 10/1000 = 0.01\$
The total current drawn from the power supply will of course be 4x that since there are 4 branches, all at 10mA.
Now, remove one branch, and what happens? Each branch still has \$I=V/R = 10/1000 = 0.01\$ since that hasn't changed at all. What has changed though is the total current. Since there are now only 3 branches the total current drops to 30mA, as that is 3 x 0.01.
The only ways the current through any branch could change is if a) the resistance in that one branch changes, or b) the voltage of the power source changes.