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luda_lava [24]
3 years ago
8

Four20 Ohm resistors are connected in parallel and the combination is connected to a 20-v emf device. The current in any one of

the resistors is: A) 0.25 A B) 4.0 A C) 5.0 A D) 100 A E) L0A
Physics
1 answer:
zloy xaker [14]3 years ago
4 0

Answer:

The current in any one of the resistors is 1 A.

Explanation:

It is given that, four 20 ohm resistors are connected in parallel. In parallel combination of resistors, the voltage across each and every resistor is same while the current divides. The equivalent resistance of all resistors is given by :

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}

\dfrac{1}{R_{eq}}=\dfrac{1}{20\ \Omega}+\dfrac{1}{20\ \Omega}+\dfrac{1}{20\ \Omega}+\dfrac{1}{20\ \Omega}

R_{eq}=5\ \Omega

Current flowing in the entire circuit can be calculated using Ohm's law as :

Current, I=\dfrac{V}{R_{eq}}

I=\dfrac{20\ V}{5\ \Omega}

I = 4 A

Since, in parallel combination current divides. So, current flowing in all four resistor divides and is 1 A. Hence, this is the required solution.

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