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2 years ago

9

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

1 answer:

kvasek [131]2 years ago

7 0

Answer:

Value of $R_1$ is 7.466 ohm

Explanation:

Let emf of the battery is e

In first case when only $R_1$ is there current through battery is 2.2 A

So $2.2=\frac{e}{R_1}$

$e=2.2R_1$ ......eqn 1

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

So $1.6=\frac{e}{R_1+2.8}$

$e=1.6R_1+4.48$ ....eqn 2

Fro eqn 1 and eqn 2

$2.2R_1=1.6R_1+4.48$

$0.6R_1=4.48$

$R_1=7.466ohm$

So value of $R_1$ is 7.466 ohm

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