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enyata [817]
3 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
Physics
1 answer:
kvasek [131]3 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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