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A loop circuit has a resistance of R1 and a current of 2.2 A. The current is reduced to 1.6 A when an additional 2.8 Ω resistor is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω

Respuesta :

Answer:

Value of [tex]R_1[/tex] is 7.466 ohm

Explanation:

Let emf of the battery is e

In first case when only [tex]R_1[/tex] is there current through battery is 2.2 A

So [tex]2.2=\frac{e}{R_1}[/tex]

[tex]e=2.2R_1[/tex]......eqn 1

In second case current is reduced to 1.6 A when additional resistance of 2.8 ohm is added in series

So [tex]1.6=\frac{e}{R_1+2.8}[/tex]

[tex]e=1.6R_1+4.48[/tex]....eqn 2

Fro eqn 1 and eqn 2

[tex]2.2R_1=1.6R_1+4.48[/tex]

[tex]0.6R_1=4.48[/tex]

[tex]R_1=7.466ohm[/tex]

So value of [tex]R_1[/tex] is 7.466 ohm

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