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rosijanka
1 month ago
15

The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time

the bulbs (resistors) stay lit in each circuit.
Physics
1 answer:
Sav [3.1K]1 month ago
3 0
This involves circuit analysis through simplification of the resistors and capacitors. We need to determine the time constant for each circuit in figures A, B, C, D, and E. This leads to ranking the duration the bulbs remain lit from longest to shortest based on each circuit's time constant. The ranking for the time constants is C > A = E > B > D. Capacitance plays a pivotal role in electrostatics, and devices called capacitors are vital components in electronic circuits. When more charge is applied to a conductor, the voltage escalates proportionately. The capacitance of a conductor is quantified as C = q/v. Adding resistors in series raises resistance while parallel configurations reduce it, conversely increasing capacitance in parallel and diminishing it in series. Thus, circuits with greater time constants take longer to discharge.
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The spring is now compressed so that the unconstrained end moves from x=0 to x=L. Using the work integral W=∫xfxiF⃗ (x⃗ )⋅dx⃗ ,
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solution:

the spring force applied by a spring with spring constant k can be expressed as

F(x)=-kx

where k acts as the spring constant

and x indicates the spring's deformation

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W=\int\limits^L_0 {} \, dW

the amount of work done by the spring when moving from x=0 to x=L

W=-kx^2/2

substituting the limits x=0 and x=L

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ANSWER

W=-kL^2/2

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "
kicyunya [3294]

Answer:

a) Q = C*emf

b)  Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let V_{R}represent the Voltage across the Resistor

V_{c}and

represent the Voltage across the capacitor

Implementing KVL;

emf - V_{R} - V_{c} = 0\\

.........................(1)

Since this is a series connection, the same current traverses through the circuit

V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C

Integrating V_{c}and V_{R}into equation (1)

emf - IR - Q/C = 0

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

I = 0A\\emf = Q/C\\Q = C* emf

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\

Substituting the values of t and I₀ into the aforementioned formula for I

I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf

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