electric circuit problems with diagrams
Answers
Answer: With having the power of 2\Omega2Ω resistor, we can find the current through it as below
P=RI^2_{2\Omega }\Rightarrow I_{2\Omega }=\sqrt{\frac{8}{2}}=2\ {\rm A}
P=RI
2Ω
2
⇒I
2Ω
=
2
8
=2 A
Before closing the switch: The resistors are in series so R_{eq}=2+4=6\ \OmegaR
eq
=2+4=6 Ω and {\mathcal E}=R_{eq}IE=R
eq
I or {\mathcal E}=Q_{tot}/C_{eq}E=Q
tot
/C
eq
so
{\mathcal E}=R_{eq}I=6\times 2=12\ {\rm V}
E=R
eq
I=6×2=12 V
And the capacitors are in series so C_{eq}=\frac{6\times 3}{6+3}=2\ \mu {\rm F}C
eq
=
6+3
6×3
=2 μF. Recall that all capacitors in series have the same charge i.e. Q_3=Q_6=Q_{tot}Q
3
=Q
6
=Q
tot
\Rightarrow Q_{tot}=C_{eq}{\mathcal E}{\rm =2\times 12=24\ }\mu {\rm C}
⇒Q
tot
=C
eq
E=2×12=24 μ C
After closing the switch: since there are no components between aa and bb, so the V_a=V_bV
a
=V
b
. The point cc and dd are directly connected to the positive and negative terminals of the battery, respectively so V_c=12{\rm \ V}V
c
=12 V and V_d=0V
d
=0.
In steady state, we only have current through resistors so again the current in the circuit is I=2{\rm \ A}I=2 A. Apply the loop rule to the adcadc to find the V_aV
a
as
\begin{gather*} V_a+2I=V_c \\ \Rightarrow V_a=12-2\left(2\right)=8\quad {\rm V} \\ \therefore V_a=V_b=8\ {\rm V}\end{gather*}
V
a
+2I=V
c
⇒V
a
=12−2(2)=8V
∴V
a
=V
b
=8 V
Now with knowing V_a,V_bV
a
,V
b
and V_cV
c
, we can find the potential drop across the capacitors and subsequently the charges.
\begin{align*}{Q'_3}&=C_3\left|V_{bc}\right|\\&=C_3\left|V_c-V_b\right|\\&=3\times (12-8)\\&=12\quad{\rm \mu C} \\\\ {Q'_6}&=C_6\left|V_{bd}\right|\\&=C_6\left|V_b-V_d\right|\\&=6\times (8-0)\\&=48\quad {\rm V}\end{align*}
Q
3
′
Q
6
′
=C
3
∣V
bc
∣
=C
3
∣V
c
−V
b
∣
=3×(12−8)
=12μC
=C
6
∣V
bd
∣
=C
6
∣V
b
−V
d
∣
=6×(8−0)
=48V
Therefore, we get
\begin{align*} \Delta Q_3&={Q'_3}-Q_3\\&=12-24\\&=-12\quad {\rm \mu C} \\\\ Q_6&={Q'_6}-Q_6\\&=48-24\\&=24\quad {\rm \mu C}\end{align*}
ΔQ
3
Q
6
=Q
3
′
−Q
3
=12−24
=−12μC
=Q
6
′
−Q
6
=48−24
=24μC
