A series circuit contains two devices, one with a resistance of 10 ohms and one with a resistance of 4 ohms. If the generator supplies a voltage of 42 V, what is the magnitude of the current?

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A series circuit contains two devices, one with a resistance of 10 ohms and one with a resistance of 4 ohms. If the generator supplies a voltage of 42 V, the magnitude of the current is 3A . R = r1+r2 , R = 10+4 =14 ohm , V=IR , 42 = I*14 , I = 42/14 =3A.

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Asked 3/6/2019 2:14:52 PM

Updated 10/30/2020 3:03:57 PM

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Cecille

A series circuit contains two devices, one with a resistance of 10 ohms and one with a resistance of 4 ohms. If the generator supplies a voltage of 42 V, the magnitude of the current is 3A .
R = r1+r2 , R = 10+4 =14 ohm , V=IR , 42 = I*14 , I = 42/14 =3A.

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A series circuit contains two devices, one with a resistance of 10 ohms and one with a resistance of 4 ohms. If the generator supplies a voltage of 42 V, what is the magnitude of the current?

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A series circuit contains two devices, one with a resistance of 10 ohms and one with a resistance of 4 ohms. If the generator supplies a voltage of 42 V, the magnitude of the current is 3 amperes. Solution: The net effective resistance of 10-ohms and 4-ohms in series is 14 ohms; I = E/R = 42V/14-ohms = 3 amperes.

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