Chapter 5: Problem 7
Capacitors \(\mathrm{A}\) and \(\mathrm{B}\) of \(0.5\) and \(0.2 \mu \mathrm{F}\), respectively, are connected in series across a \(140 \mathrm{~V}\) d.c. supply. What are the charges and voltages for each capacitor? Repeat with \(\mathrm{A}\) and \(\mathrm{B}\) in parallel.
Short Answer
Expert verified
A. In series: Capacitor A - 20 μC, 40 V; Capacitor B - 20 μC, 100 V
B. In parallel: Capacitor A - 70 μC, 140 V; Capacitor B - 28 μC, 140 V
Step by step solution
01
Calculate equivalent capacitance of series connection
The equivalent capacitance of capacitors connected in series is given by the formula:
\(1/C_{eq} = 1/C_A + 1/C_B\)
We are given,
\(C_A = 0.5\, \mu F\) and \(C_B = 0.2\, \mu F\)
Calculating \(C_{eq}\) for series connection:
\(1/C_{eq} = 1/0.5 + 1/0.2\)
So, \(C_{eq} = 1 / (1/0.5 + 1/0.2) \approx 0.1429 \,\mu F\)
02
Calculate charge on each capacitor in series connection
For capacitors connected in series, the charge on each capacitor is the same.
Using the formula \(Q = C_{eq} * V\),
Charge in series connection, \(Q_s = 0.1429\, \mu F * 140\, V \approx 20 \, \mu C\)
03
Calculate the voltage across each capacitor in series connection
Using the formula \(V = Q/C\),
\(V_A = Q_s / C_A = 20\, \mu C / 0.5\, \mu F = 40\, V\)
\(V_B = Q_s / C_B = 20\, \mu C / 0.2\, \mu F = 100\, V\)
Result: In the series connection, both capacitors have a charge of \(20\, \mu C\). Capacitor A has a voltage of \(40\, V\) and Capacitor B has a voltage of \(100\, V\).
04
Calculate equivalent capacitance of parallel connection
The equivalent capacitance of capacitors connected in parallel is given by the formula:
\(C_{eq} = C_A + C_B\)
Calculating \(C_{eq}\) for parallel connection:
\(C_{eq} = 0.5\, \mu F + 0.2\, \mu F = 0.7\, \mu F\)
05
Calculate total charge stored in parallel connection
For capacitors connected in parallel, the total charge stored is given by:
\(Q_{total} = C_{eq} * V\)
Charge in parallel connection, \(Q_{total} = 0.7\, \mu F * 140\, V = 98\, \mu C\)
06
Calculate the voltage across each capacitor in parallel connection
For capacitors connected in parallel, the voltage across each capacitor is the same as the supply voltage.
\(V_A = V_B = 140\, V\)
07
Calculate the charge on each capacitor in parallel connection
Using the formula \(Q = C * V\),
\(Q_A = C_A * V_A = 0.5\, \mu F * 140\, V = 70\, \mu C\)
\(Q_B = C_B * V_B = 0.2\, \mu F * 140\, V = 28\, \mu C\)
Result: In the parallel connection, Capacitor A has a charge of \(70\, \mu C\) and a voltage of \(140\, V\), while Capacitor B has a charge of \(28\, \mu C\) and a voltage of \(140\, V\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equivalent Capacitance
Understanding how to combine capacitors is a fundamental skill in circuit analysis. Equivalent capacitance is a way of simplifying complex circuits with multiple capacitors into a single capacitor that has the same effect on the circuit as the original combination.
In series arrangements, capacitors' values add inversely, leading to a lower equivalent capacitance given by the formula \(1/C_{eq} = 1/C_1 + 1/C_2 + ... + 1/C_n\). This can be understood by imagining each capacitor as a bottle neck for charge flow; more bottle necks in series mean less overall flow and thus a smaller equivalent capacity. Conversely, in parallel arrangements, capacitors simply add up directly, mirroring the idea that parallel paths allow for more charge storage; hence, \(C_{eq} = C_1 + C_2 + ... + C_n\).
Our exercise illustrates this by first finding a lower equivalent capacitance for series-connected capacitors (0.1429 \(\mu F\)) and a higher one for parallel-connected capacitors (0.7 \(\mu F\)).
In series arrangements, capacitors' values add inversely, leading to a lower equivalent capacitance given by the formula \(1/C_{eq} = 1/C_1 + 1/C_2 + ... + 1/C_n\). This can be understood by imagining each capacitor as a bottle neck for charge flow; more bottle necks in series mean less overall flow and thus a smaller equivalent capacity. Conversely, in parallel arrangements, capacitors simply add up directly, mirroring the idea that parallel paths allow for more charge storage; hence, \(C_{eq} = C_1 + C_2 + ... + C_n\).
Our exercise illustrates this by first finding a lower equivalent capacitance for series-connected capacitors (0.1429 \(\mu F\)) and a higher one for parallel-connected capacitors (0.7 \(\mu F\)).
Series and Parallel Capacitors
Delving deeper into capacitor arrangements, we can distinguish between their behavior in series and parallel connections. For series circuits, recall that the charge \(Q\) remains the same across all capacitors. This is akin to saying that the amount of electrical fluid (charge) is conserved as it flows through each 'bottle'.
In contrast, for capacitors in parallel, it's the voltage across them that's identical, as each capacitor is directly connected to the voltage source, similar to different paths in a park all showing the same scenery.
The calculated charges in our exercise illustrate this: in series, the charge is uniformly 20 \(\mu C\) on both capacitors. In parallel, however, each capacitor stores a charge propotional to its capacitance (70 \(\mu C\) for Capacitor A and 28 \(\mu C\) for Capacitor B), but at the same voltage (140V).
In contrast, for capacitors in parallel, it's the voltage across them that's identical, as each capacitor is directly connected to the voltage source, similar to different paths in a park all showing the same scenery.
The calculated charges in our exercise illustrate this: in series, the charge is uniformly 20 \(\mu C\) on both capacitors. In parallel, however, each capacitor stores a charge propotional to its capacitance (70 \(\mu C\) for Capacitor A and 28 \(\mu C\) for Capacitor B), but at the same voltage (140V).
Capacitor Charge and Voltage
The relationship between capacitor charge and voltage is given by \(Q = CV\), where \(Q\) is the charge in coulombs, \(C\) is the capacitance in farads, and \(V\) is the voltage in volts. This equation encapsulates one of the fundamental characteristics of a capacitor: its ability to store energy in the form of electrical charge.
Our exercise showcases this beautifully, With the parallel combination yielding the same voltage for both capacitors but different charges reflective of their individual capacitances.
Capacitor in Series:
When connected in series, the voltage across each capacitor differs based on its capacitance, as seen with our 40V across Capacitor A and 100V across Capacitor B, despite them sharing the same charge.Capacitor in Parallel:
In parallel configurations, each capacitor has the same voltage as the source but holds a different charge relative to its capacitance. The exquisite balance of these equations allows for the precise tuning of electric circuits, tailored just as needed in engineering designs.Our exercise showcases this beautifully, With the parallel combination yielding the same voltage for both capacitors but different charges reflective of their individual capacitances.