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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?
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♦ Relevant knowledgeThe circuit with capacitors and resistors linked with the source of voltage is referred to as an RC-circuit. When the switch is shut in a circuit like this the transition process begins until all capacitors are fully charged and there is no flow of current in the branches that contain them. The amount of time it takes the capacitors to achieve 63 percent of their full charge is known as”the time-constant” of circuit.
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]
I is the current, and R the 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]
C here 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