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Sidana [21]
2 years ago
10

What is the definition of electrical power?

Physics
1 answer:
Nonamiya [84]2 years ago
6 0
An electric power measure the rate of electrical energy transfer by an electric circuit per unit of time.
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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 "
blsea [12.9K]

Answer:

a) Q = C*emf

b)  Reduction 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 connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let V_{R} = Voltage dropped across the Resistor

V_{c} = Voltage dropped across the capacitor

Applying KVL;

emf - V_{R}  - V_{c} = 0\\.........................(1)

Since the connection is in series, the same current flow through the circuit

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

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

emf - IR - Q/C = 0

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

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

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

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

Putting the values of t and I₀ into the formula for I written above

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

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

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

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