Question

A 6.00 volt battery is in a circuit with two resistors, A and B, connected in parallel. If the current flowing

out of the battery is 90.0 mA, what is the value in ohms of B if A is 100 ?

Answer 1

Answer:
`200\Omega`

Step-by-step explanation:

Given

The battery used is 6V in magnitude

The current flow is 90 mA

If the resistance of A is
`100\Omega`

Net resistance of the circuit is

`R_(net)=(V)/(I)\\\\\Rightarrow R_(net)=(6)/(90* 10^(-3))\\\\\Rightarrow R_(net)=(200)/(3)\Omega`

If A and B are in parallel then their net resistance must be
`(200)/(3)`

`(1)/((200)/(3))=(1)/(100)+(1)/(R_b)\\\\\Rightarrow (1)/(R_b)=(3)/(200)-(1)/(100)\\\\\Rightarrow (1)/(R_b)=(3-2)/(200)\\\\\Rightarrow R_b=200\Omega`