Consider the circuit shown in (Figure 1) . The battery has emf 92.0 V and neglig
ID: 1553376 • Letter: C
Question
Consider the circuit shown in (Figure 1) . The battery has emf 92.0 V and negligible internal resistance. R2 = 2.00 , C1 = 4.00 F , and C2 = 8.00 F .After the capacitors have attained their final charges, the charge on C1 is Q1 = 18.0 C .
Problem 25.72 Consider the circuit shown in (Figure 1). The battery has emf 92.0 V and negligible internal resistance R2 32.00 s ,C1 400 HF and C2 8.00 uF After the capacitors have attained their final charges, the charge on 01 is 01 18 Auc of 1 Figure 1 Rs Part A What is the final charge on C2? Express your answer with the appropriate units HA Value Units Submit My Answers Give Up Part B What is the resistance R1? Express your answer with the appropriate units R1 Value Units Submit My Answers Give Up Provide Feedback ContinueExplanation / Answer
Concept used : Relation between potential drop(V) across capacitor , Capacitance of capacitor(C) and charge on capacitor (Q) is
CV = Q
Part A
Potential drop across C1= V1 = Q1/C1 = 18/4 = 4.5 V
Potential drop across C2 = V2 = Potential drop across C1 =4.5 V (C1 and C2 parallel )
So charge on C2 = Q2 =C2V2 = 8 * 4.5 = 36 uC
Part B
Potential drop across R2 =V = potential drop across C1 = 4.5 ( they are in parallel )
When capacitors are fully charged then no current flows in paths containing capacitors so
current through R1= current through R2=I = V/R2= 4.5/2 = 2.25 A
Potential drop across battery = 92 V
Out of this 4.5 V dropped across R2 so remaining i.e. 92 - 4.5 = 87.5 V will drop across R1
Hence potential drop across R1 = V' =87.5 V
So R1 = V'/I = 87.5/2.25 = 38.89 ohm