Question

A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected from the battery. A resistor of resistance R = 185 Ω is connected across the capacitor, and the capacitor discharges through the resistor.
What is the charge q on the capacitor when the current in the resistor is 0.180 A?

Answer 1

Answer:

1.99×10^-4coulombs

Explanation:

The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV

If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.

We need to calculate the voltage on the resistor first when 0.18A current is passed through it.

V = 0.18 × 185

V = 33.3Volts

The charge Q on the resistor will be;

Q = CV

Were C = 6.00 μF, V = 33.3

Q= 6×10^-6 ×33.3

Q = 0.0001998

Q= 1.99×10^-4Coulombs

Answer 2

Given the current in the resistor, the charge q on the capacitor is 1.988 × 10⁻⁴C

Given the data in the question;

• Capacitance of the Capacitor; $C = 6.00 uF = 6*10^{-6}F$

• Electromotive force; $emf = 50.0 V$

• Resistance; $R = 185 ohms$

• Current in the resistor; $I = 0.180A$

• Charge on the capacitor; $q =\ ?$

First, we calculate the voltage. From Ohm's Law:

$V =IR$

Where $V$ is the voltage, $I$ is the Current and $R$ is the resistance

We substitute in our values

$V = 0.180A\ *\ 185ohms\\\\V = 33.3Volts$

Using the equation for  measuring the capacitance (C) of a capacitor:

$C = \frac{q}{V}$

Where q is the the charge in the capacitor, and V is the Voltage

We substitute in the equation

$6*10^{-6}F = \frac{q}{33.3Volts}\\\\q = 6*10^{-6}F\ *\ 33.3Volts\\\\q = 1.998*10^{-4}C$

Therefore, Given the current in the resistor, the charge q on the capacitor is 1.988 × 10⁻⁴C

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