Question

. A student has three resistances R1 = 3Ω, R2 = 6Ω and an unknown resistance R3. She made a circuit

out of these three, which is shown below. The current she obtained in the entire circuit was found by her

to be 5A. Calculate the unknown resistance R3.​

Answer 1

Solution:

R1=3 ohms

Second resistance=R2 ohms

As they are connected in series :

Eff R= R1+R2

=3+R2 ohms

V=12V

V2=6V

AS IT IS SERIES CONNECTION:

V=V1+V2

12= V1+6

V1= 6VOLTS

POTENTIAL DROP ACROSS FIRST RESISTANCE IS 6V

By ohms law:

V1=IR1

6=Ix3

I=6/2=3A

In series connection:

I=I1=I2

So current ACROSS unknown resistance is also 2A

V2=IR2

6=2xR2

R2=6/2=3ohms

Unknown resistance =3ohms

Effective resistance in Series = 3+3=6ohms

PLS MARK BRAINLIEST