(b) Two resistors, when connected in series have total resistance of 25 ohms. If they are
connected in parallel the value goes down to 6 ohms. Find their values.
Answers
Answer: The total resistance of resistors in series is equal to the sum of their individual resistances. That is, R
total
=R
1
+R
2
+R
3
.
The total resistance will always be less than the value of the smallest resistance That is,
R
total
1
=
R
1
1
+
R
2
1
+
R
3
1
thatis,R=R
total
−1
.
In this case, the total resistance of the resistors connected in series is given as R
1
+R
2
=45 - eqn 1
The total resistance of the resistors connected in parallel is given as
R
1
1
+
R
2
1
=
10
1
thatis,
R
1
1
+
R
2
1
=0.1 - eqn 2
Putting the value of R
1
+R
2
in eqn 2, we get, R
1
R
2
=10×45=450ohms.
Now, From the identity, (R
1
−R
2
)
2
=(R
1
+R
2
)
2
−4R
1
R
2
,weget,R
1
−R
2
=15ohms - eqn 3.
Adding eqn 2 and eqn 3,
2R
1
=60ohmsThatis,R
1
=30ohms.
Substituting this value in eqn 3, we get, R
2
=(30−15)=15ohms.
Hence, the values of the resistors are 30 ohms and 15 ohms
Answer:
How would you find them? Well, you have a system of two equations and two unknowns R1 and R2. That means that there is a unique solution. All you have to do is setup those two equations and solve the system using algebra. The two equations come from knowledge of how resistances combine when they are connected in series, and how they combine when they are connected in parallel
Explanation: