The charge accumulated on the capacitor connected in the following circuit is ______ $$\mu$$C. (Given $$C = 150 \; \mu$$F)

step 1: steady state
capacitor is open → no current through AB

So the diamond reduces to two parallel branches between top and bottom nodes

• right branch: 6Ω + 4Ω = 10Ω
• left branch: 1Ω + 2Ω = 3Ω

step 2: equivalent of parallel

$$R_{eq}=10∥3=\frac{30}{13}Ω$$

Total current from 10 V source:

$$I=\frac{10}{30/13}=\frac{130}{30}=\frac{13}{3}\text{ A}$$

step 3: current division

current in left branch (1Ω + 2Ω):

$$I_L=I\cdot\frac{10}{10+3}=\frac{13}{3}\cdot\frac{10}{13}=\frac{10}{3}\text{ A}$$

current in right branch:

IR=133−103=1 AI_R = \frac{13}{3} - \frac{10}{3} = 1 \text{ A}IR​=313​−310​=1 A

step 4: find potentials

take bottom node = 0 V, top = 10 V

Point A (on left branch):

drop across 1Ω:

$$V_A=10-\frac{10}{3}=\frac{20}{3}$$

Point B (on right branch):

drop across 6Ω:

$$V_B=10-6(1)=4$$

step 5: capacitor voltage

$$V_{AB}=\frac{20}{3}-4=\frac{8}{3}$$

step 6: charge

$$Q=CV=150\times\frac{8}{3}=400μC$$