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Galina-37 [17]
2 years ago
12

The lamp has a resistance of 10 ohms each resistor has a resistance of 10 ohms what is the total resistance of the circuit

Physics
1 answer:
prisoha [69]2 years ago
7 0
Mark Brainliest please

Answer : if connected series, 20 ohms
And if connected parallel, answer will be less than 20 ohms

When resistors are wired in series, the total circuit resistance increases because each resistor contributes opposition to the circuit's current flow. Therefore if a 10 ohm resistor is placed in series with another 10 ohm resistor, the total resistance contributed by the two resistors is 20 ohms.
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To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p
il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

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Explanation:

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At the end of one full time period, the ant has returned to where it was at the beginning of the time period. Its displacement is <em>zero</em>.

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