Solveeit Logo

Question

Question: What is total resistance , between A and B in the circuit shown in the figure. ![](https://www.ved...

What is total resistance , between A and B in the circuit shown in the figure.

A. 1Ω1\Omega
B. 0.5Ω0.5\Omega
C. 2Ω2\Omega
D. 3Ω3\Omega

Explanation

Solution

Clearly, the given figure consists of a combination of resistances which are connected in parallel and series with each other. In a series connection, we know that the flow of current through the circuit is a constant . While in parallel circuit, we know that the flow of voltage through the circuit is a constant. Using the formula, we can find the total resistance.

Formula used:
Rs=R1+R2R_{s}=R_{1}+R_{2} and 1Rp=1R1+1R2\dfrac{1}{R_{p}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}

Complete step-by-step answer:
We know that Ohm’s law gives the relationship between voltage VV, current II and resistance RR. It is given as V=IRV=IR at constant temperature .
We also know that, since current in the series connection is constant or same amount of current flows through the circuit , then the effective resistance RsR_{s}is given by Rs=R1+R2R_{s}=R_{1}+R_{2}
Also, since voltage in the parallel connection is constant same amount of voltage flows through the circuit , then the effective resistance RpR_{p} I given by 1Rp=1R1+1R2\dfrac{1}{R_{p}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}
Clearly from the above figure, AB is in series with BC, then we have,
R=4+2=6ΩR=4+2=6\Omega
Again, ABC is in parallel with AC,
1R=14+16=312+212=512\dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{3}{12}+\dfrac{2}{12}=\dfrac{5}{12}
    R=125\implies R=\dfrac{12}{5}
This new AC is in series with CD
R=125+2=12+105=225R=\dfrac{12}{5}+2=\dfrac{12+10}{5}=\dfrac{22}{5}
Again, ACD is in parallel with AD,
1R=522+14=20+2288=4488=12\dfrac{1}{R}=\dfrac{5}{22}+\dfrac{1}{4}=\dfrac{20+22}{88}=\dfrac{44}{88}=\dfrac{1}{2}
    R=2\implies R=2
Again, AD is in series with DE
R=2+2=4R=2+2=4
Again ADE is in parallel with AE
1R=14+14=12\dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}
    R=2\implies R=2
Again AE is in series with EF
R=2+2=4R=2+2=4
AEF is in parallel with AF
1R=14+14=24=12\dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}
    R=2\implies R=2

So, the correct answer is “Option C”.

Note: Since it is not possible, every time to calculate the current or voltage passing through the circuit. Another, easy way to identify if the resistances are in series or parallel connections is by observing the circuit diagram. If two or more resistances have a pair of common points then, then they are said to be in parallel circuit. If two or more resistances have less than two or only one common point then, they are said to be in series circuit.