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Naily [24]
3 years ago
12

Five 60 ohm resistors are connected in parallel. What is their equivalent resistance?

Physics
2 answers:
Mnenie [13.5K]3 years ago
7 0

Answer:

i think the answer is 12 ohms

plz mark me as brainliest :)

algol133 years ago
4 0

Answer:

12 ohms

Explanation:

The five 60 Ohms resistance can be represented as follows:

R1 = 60 Ohms

R2 = 60 Ohms

R3= 60 Ohms

R4 = 60 Ohms

R5 = 60 Ohms

Rt =?

Since they are in parallel connections, the equivalent resistance (Rt) can be calculated as follows :

1/Rt = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5

1/Rt = 1/60 + 1/60 + 1/60 + 1/60 + 1/60

1/Rt = ( 1+1+1+1+1)/60

1/Rt = 5/60

1/Rt = 1/12

Cross multiply to express in linear form

Rt = 12 ohms

Therefore the equivalent resistance is 12ohms

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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
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Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0  x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12    (pico 'p' = 10^-12)

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the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))  

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR)  --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides  

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x 10^{-9  

t = 23.9ns

Thus, it took 23.9ns  for the potential difference between the deflection plates to reach 75 volts

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