Resistance in individual branches of parallel systems vs. resistance of series circuits

A straightforward circuit with a parallel system component activated by a switch in front of resistor "C":
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so I have presented with a certain circuit with a parallel set within, which is activated or connected by the switch shown (in this circuit, A, B and C all have an equivalent resistance measurement).

So, before the switch is closed, the total resistance of the circuit can be calculated in series, with respect to resistors A and B, which should be $2R$ ($R$ being the value of the resistance of each resistor), when the switch is closed and the parallel set is connected, the total resistance of the circuit decreases and the current going into the battery should increase; now the total resistance should be $R + \frac{R}{2}$, or $\frac{3R}{2}$, since the whole parallel set can be treated as one large resistor in series with respect to A,

So how does the voltage and current through resistor B compare to when it connected in series with A vs. when the parallel system is activated with the switch, adding resistor “C” ?

• What is your question? I don't see one question mark in this wall of text. – Alfred Centauri May 24 '18 at 2:59
• How does the resistance, voltage and current through resistor B compare to when it is connected in series with “A” versus when the switch is closed and it becomes part of the parallel system with “C”? – Jack Scrugggs May 24 '18 at 3:41
• Jack, you need to copy the text in your comment and add it as an edit to your question. – Chappo May 24 '18 at 7:37
• Jack, I'm not sure what you mean by "resistance through resistor B" but it seems to me that you have all you need to solve for the voltage across and current through resistor B when the switch is open and when the switch is closed. So, when you solve for the two cases, what do you find? – Alfred Centauri May 24 '18 at 12:22
• When the parallel system is added, does the current through A increase or stay the same, even though another branch is added – Jack Scrugggs May 24 '18 at 13:21