A parallel circuit has two 8.0-ohm resistors and a power source of 9.0 volts. If a 12.5-ohm resistor is added to the circuit in parallel, how will the current be affected and what value will it have?

Question

anonymous · Answer

There will be a decrease in the effective resistance (total resistance), so the current will increase.

anonymous · Answer

A) current will increase; 2.97 amps   B) current wil increase; .55 amps    C) current will decrease 2.97 amps    D) Current will decrease .55 amps        But how do I know how much it will increase? @Abidlumsden

anonymous · Answer

The answer is A. The resistance after the 12.5 ohm resistor is added, is equal to 3 ohm. $$1\div R = 1\div8+1\div8+1\div12.5$$ Where "R" is the total resistance. 

The voltage is 9 volts, to find the current you divide the voltage by the total resistance. 
Since the resistance decreased, and the voltage stayed the same, the current would increase.

anonymous · Answer

Thank you so much!! @Abidlumsden

anonymous · Answer

No prob :)

anonymous · Answer

The effective resistance is easy to calculate:$$\frac{ 1 }{ R _{eff} }=\frac{ 1 }{ R _{1} }+\frac{ 1 }{ R _{2}}...+\frac{ 1 }{ R _{n} }$$where Reff is the effective resistance; R1 is the first resistor in parallel; R2 is the second resistor in parallel; Rn is the nth resistor in  parallel; and n is the number of resistors.  

Finding the current is as simple as solving Ohm's law:$$I=\frac{ V }{ R _{eff} }$$