Question

In fig. the equivalent resistance of the following combination of resistors $r_1,r_2,r_3$ and $r_4$ if $r_1=r_2=r_3=r_4=2.0\Omega$, between the points A and B is

• A. $2.0\Omega$
• B. $4.0\Omega$
• C. $6.0\Omega$
• D. $5.0\Omega$

Answer 1

The correct option is D $5.0\Omega$

In this case, the first two resistors of 2 ohms each are connected in series and the third and fourth resistors of 2 ohms each are connected in parallel. The first set is connected in series with the second set of resistors.
The resistances R3 and R4 that are connected in parallel have a resistance given as $\frac{1}{2}+\frac{1}{2}=\frac{1}{1}=1ohm$. That is, these two resistances is reduced to a single resistance of 1 ohm.
The resistances R1 and R2 that are connected in series have a resistance given as 2+2 = 4 ohms.
This combined resistance when connected in series with reduced resistance of 1 ohm gives a total resistance of 4+1 = 5 ohms.
Hence, the total resistance in the circuit is 5 ohms.