Question

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

Answer 1

Value of
`R_1` is 7.466 ohm

Step-by-step 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=(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=(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