Ishani and John now try a problem involving a charging capacitor.An uncharged capacitor with C = 3.62 𝜇F and a resistor with R = 9.0 105 Ω are connected in series to a 12.0 V battery. The switch is closed at time t = 0, allowing the capacitor to charge.Find the maximum charge on the capacitor and the maximum current through the resistor.

The maximum charge (Q) on the capacitor can be found using the formula Q = CV, where C is the capacitance and V is the voltage.

Step 1: Convert the capacitance from microfarads to farads. 1 microfarad (uF) is equal to 1e-6 farads (F). So, C = 3.62 uF = 3.62e-6 F.

Step 2: Substitute the values of C and V into the formula. V = 12.0 V.

Q = CV Q = (3.62e-6 F)(12.0 V) Q = 43.44e-6 C

So, the maximum charge on the capacitor is 43.44 microcoulombs (uC).

The maximum current (I) through the resistor can be found using Ohm's law, I = V/R, where V is the voltage and R is the resistance.

Step 3: Convert the resistance from ohms to megaohms. 1 megaohm (MOhm) is equal to 1e6 ohms. So, R = 9.0 MOhm = 9.0e6 Ohm.

Step 4: Substitute the values of V and R into the formula.

I = V/R I = (12.0 V)/(9.0e6 Ohm) I = 1.33e-6 A

So, the maximum current through the resistor is 1.33 microamperes (uA).

The maximum charge (Q) on the capacitor can be found using the formula Q = CV, where C is the capacitance and V is the voltage.

So, Q = 3.62 μF * 12.0 V = 43.44 μC.

The maximum current (I) through the resistor can be found using Ohm's law, I = V/R.

So, I = 12.0 V / 9.0 * 10^5 Ω = 1.33 * 10^-4 A.

So, the maximum charge on the capacitor is 43.44 μC and the maximum current through the resistor is 1.33 * 10^-4 A.

The maximum charge (Qmax) on the capacitor can be found using the formula:

Qmax = CV

where C is the capacitance and V is the voltage. Substituting the given values:

Qmax = 3.62 μF * 12.0 V = 43.44 μC

The maximum current (Imax) through the resistor can be found using Ohm's law:

I = V/R

where V is the voltage and R is the resistance. Substituting the given values:

Imax = 12.0 V / 9.0 * 10^5 Ω = 1.33 * 10^-4 A

So, the maximum charge on the capacitor is 43.44 μC and the maximum current through the resistor is 1.33 * 10^-4 A.