If a 3 Ω and a 1.5 Ω resistor are wired in parallel and the combination is wired in series to a 4 Ω and a 10 V emf device then the current in the 3 Q resistor is 0.667 Ampere
<h3>What is resistance?</h3>
Resistance is the obstruction of electrons in an electrically conducting material.
The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.
V=IR
As given in the problem 3 Ω and a 1.5 Ω resistor are wired in parallel
Their equivalent resistance would be
1/Re= 1/R1 + 1/R2
1/Re = 1/3 +1/1.5
Re= 1 Ω
Now this parallel equivalent is connected with a series combination with a 4Ω resistor
For calculating equivalent resistance in series combination.
R = R1 + R2
R = 1 + 4
R = 5 Ω
As the 10 V emf device is connected then the current flowing can be calculated by using Ohms law
V=IR
10 = 5×I
I = 2 ampere
as the voltage is divided between parallel resistor combination and series resistor
The voltage drop across parallel combination = 10×(1/1+4) volt
= 2 volt
Now the current flows through a 3 Ω resistor by using Ohms law
I = V/R
I= 2/3
I = 0.667 Ampere
Thus, the current flowing through the 3 Ω resistor comes out to be 0.667 Ampere
Learn more about resistance from here
brainly.com/question/14547003
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