In the circuit shown in figure C1:C2:C3 =1:2:3 . When switch is at pos...
**Solution:**
Given: C1:C2:C3 = 1:2:3
Let's assume the capacitance of C1, C2, and C3 as C1 = C, C2 = 2C, and C3 = 3C respectively.
When the switch is at position one for a long time, C1 acquires a charge Q1.
**Charging of Capacitor in Position 1:**
When the switch is in position one, C1 is connected to the voltage source. Let's assume the potential difference across C1 is V1.
Using the formula for the charge stored in a capacitor:
Q = CV
The charge stored in C1 can be calculated as:
Q1 = C1 * V1
Since Q1 is the charge acquired by C1, the same charge is supplied by the voltage source.
**Change in the Circuit Configuration:**
Now, the switch is thrown to position 2.
**Charging of Capacitors in Position 2:**
In position 2, C1 and C2 are connected in parallel to the voltage source, and C3 is disconnected.
The potential difference across C1 and C2 in position 2 is V2.
The charge stored in C1 and C2 can be calculated as:
Q1 = C1 * V2
Q2 = C2 * V2
Since C1 and C2 are connected in parallel, the total charge stored in the combination of C1 and C2 is the sum of the charges stored in each capacitor.
Q1 + Q2 = C1 * V2 + C2 * V2
Substituting the values of C1 and C2, we get:
Q1 + Q2 = (C * V2) + (2C * V2)
Q1 + Q2 = 3C * V2
Since the total charge supplied by the voltage source remains the same, we have:
Q1 = Q2 + Q3
Substituting the values of Q1 and Q2, we get:
C1 * V1 = 3C * V2
**Calculating the Ratio Q1/Q2:**
Dividing both sides of the equation by C * V2, we get:
Q1/Q2 = (3C * V2) / (C * V2)
Q1/Q2 = 3
Therefore, the ratio Q1/Q2 is 3.