In Fig. 25-37, V = 12 V, C1 = 9.8 µF, and C2 = C3 = 18 µF. Switch S is first thrown to the left side until capacitor 1 reaches equilibrium. Then the switch is thrown to the right. When equilibrium is again reached, how much charge in Coulombs is on capacitor 1?

To find the charge in Coulombs on capacitor 1, we need to use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.

In this circuit, capacitor 1 is initially charged to some voltage, let's call it V1. When switch S is thrown to the left, capacitor 1 is disconnected from the circuit and it will start discharging through the circuit. At equilibrium, the voltage across capacitor 1 is 0V.

So, the initial charge on capacitor 1 can be found using the formula Q1 = C1 * V1.

Next, when switch S is thrown to the right, capacitor 1 is connected to capacitors 2 and 3 in parallel. At equilibrium, the voltage across capacitors 2 and 3 will be equal to V.

Using the formula Q = CV, the charge on capacitors 2 and 3 at equilibrium is Q2 = C2 * V and Q3 = C3 * V.

Since capacitors 2 and 3 are in parallel, the total charge on the combination of capacitors 2 and 3 is the sum of the individual charges, Q2 + Q3.

Finally, the charge on capacitor 1 at equilibrium can be found by subtracting the charge on the combination of capacitors 2 and 3 from the initial charge on capacitor 1, Q1.

So, the charge on capacitor 1 at equilibrium is Q1 - (Q2 + Q3).

By plugging in the given values for V, C1, C2, and C3, you can calculate the charge on capacitor 1 using the equations mentioned above.