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The capacitor in the figure shown is initially uncharged. The switch is closed at t=0.

a) Immediately after the switch is closed, what is the current through each resistor? b) What is the final charge on the capacitor?

To solve the problem, you will need to understand Ohm’s law, resistance in series, resistance in parallel combination and voltage divider rule.

Calculate the equivalent resistance first by using resistances in series and parallel combinations. Next, use Ohm’s law for each resistor to calculate its current.

Use the voltage divider rule for calculation of the voltage across the capacitor. Next, use the expression to calculate the charge stored in the capacitor.

This is the Ohm’s Law expression:

[katex]V = IR[/katex]

Iis the current, andRthe resistor.This is the equivalent resistance of resistors that are connected in series.

[katex]R = {R_1} + {R_2}[/katex]

Here, [katex]{R_1}[/katex]

This is the equivalent resistance to resistors that are connected in parallel.

[katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]

The equation for the amount of charge on the capacitor is given as follows:

[katex]q = CV[/katex]

Chere is the capacitance.(a)

Let’s say that the capacitor is cut.

This is the equivalent resistance to resistors that are connected in parallel.

[katex]R = \frac{{{R_2}{R_3}}}{{{R_2} + {R_3}}}[/katex]

Substitute [katex]6.00{\rm{ }}\Omega [/katex]

[katex]\begin{array}{c}\\R = \frac{{\left( {6.00{\rm{ }}\Omega } \right)\left( {3.00{\rm{ }}\Omega } \right)}}{{6.00{\rm{ }}\Omega + 3.00{\rm{ }}\Omega }}\\\\ = 2.00{\rm{ }}\Omega \\\end{array}[/katex]

This is the equivalent resistance of resistors that are connected in series.

[katex]R’ = R + {R_1}[/katex]

Substitute [katex]8.00{\rm{ }}\Omega [/katex]

[katex]R = {R_1} + {R_2}[/katex]0

(a.1)

Ohm’s law states that the current expression is given by

[katex]R = {R_1} + {R_2}[/katex]1

Here, [katex]R = {R_1} + {R_2}[/katex]2

Substitute 42.0 V to [katex]R = {R_1} + {R_2}[/katex]2

[katex]R = {R_1} + {R_2}[/katex]4

(a.2)

Now, let’s use the ohms law to calculate the voltage drop across resistor [katex]{R_1}[/katex].

[katex]R = {R_1} + {R_2}[/katex]6

Substitute [katex]8.00{\rm{ }}\Omega [/katex]

[katex]R = {R_1} + {R_2}[/katex]8

The difference between resistors [katex]R = {R_1} + {R_2}[/katex]9

[katex]{R_1}[/katex]0

Substitute 42.0 V to [katex]R = {R_1} + {R_2}[/katex]2

[katex]{R_1}[/katex]2

The expression for the current across [katex]R = {R_1} + {R_2}[/katex]9

[katex]{R_1}[/katex]4

Substitute [katex]{R_1}[/katex]5

[katex]{R_1}[/katex]6

(a.3)

The expression for the current across [katex]{R_1}[/katex]7

[katex]{R_1}[/katex]8

Substitute [katex]{R_1}[/katex]5

[katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]0

(b)

The equation for the amount of charge on the capacitor is given as follows:

[katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]1

Here, [katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]2

The voltage across the capacitor, according to the Voltage Divider rule, is determined by

[katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]3

Substitute [katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]4

[katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]5

Substitute [katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]6

[katex]\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}[/katex]7Ans Part a.1

The current through resistor [katex]{R_1}[/katex]Part A.2

The current through resistor [katex]R = {R_1} + {R_2}[/katex]9Part A.2

The current through resistor [katex]{R_1}[/katex]7Part

The capacitor’s charge is [katex]q = CV[/katex]1