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2 years ago

Why it is advise to clean the ends of connecting wires before connecting them?

Physics

2 answers:

densk [106]2 years ago

8 0

To avoid poor contact and consequently the increase of the electrical resistance at the junction.

Svetradugi [14.3K]2 years ago

5 0

If dirt and grease were good conductors of electrical current, then we could make wire
out of dirt and grease instead of expensive copper. Sadly, they're not. So a coating of
dirt and grease on the wire can have a substantial impact on the connection, and can
even block the flow of current across the connection completely. Moreover, in the case
where the ends of the wires are to be soldered, solder does not adhere to dirty wire.

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3 years ago

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)

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