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Question: Find the value of \({R_{net}}\) between \(A\) and \(B\). ![](https://www.vedantu.com/question-sets...

Find the value of Rnet{R_{net}} between AA and BB.

A) 4040
B) 6060
C) 7070
D) 2020

Explanation

Solution

First use the series combination of R1,R2,R3{R_1},{R_2},{R_3} to find the values of Rs1{R_{s1}}. Now these Rs1{R_{s1}} will be become parallel combination with R4{R_4} .Then calculate the value of Rp1{R_{p1}} . after that R5,R6{R_5},{R_6} and Rp1{R_{p1}} are in series combination, So calculate the value of Rs2{R_{s2}} .Now R7,Rs2{R_7},{R_{s2}} are in parallel combination, Therefore we have to find the value of Rp2{R_{p2}} now R8,R9{R_8},{R_9} and Rp2{R_{p2}} are in series combination than after we calculate Rnet{R_{net}}.

Complete step by step solution:
Let us consider the following diagram

First, we must calculate Rs1{R_{s1}}
In this question, the given that,
Resistance
R1=20 R2=10 R3=10  {R_1} = 20 \\\ {R_2} = 10 \\\ {R_3} = 10 \\\

Since these resistances are in series combination, So we will use the series combination formula that is,
Rs1=R1+R2+R3\Rightarrow {R_{s1}} = {R_1} + {R_2} + {R_3}

On putting the value of these resistance we get,
Rs1=20+10+10\Rightarrow {R_{s1}} = 20 + 10 + 10
After calculation we will get,
Rs1=40\Rightarrow {R_{s1}} = 40

Now, Rs1{R_{s1}} will become a parallel combination with R4{R_4}.

We know that parallel combination formula is
1Rp1=1R4+1Rs1\Rightarrow \dfrac{1}{{{R_{p1}}}} = \dfrac{1}{{{R_4}}} + \dfrac{1}{{{R_{s1}}}}

On putting the value of these resistance we get,
1Rp1=140+140\Rightarrow \dfrac{1}{{{R_{p1}}}} = \dfrac{1}{{40}} + \dfrac{1}{{40}}
After calculation we will get,
1Rp1=240\Rightarrow \dfrac{1}{{{R_{p1}}}} = \dfrac{2}{{40}}
1Rp1=120\Rightarrow \dfrac{1}{{{R_{p1}}}} = \dfrac{1}{{20}}
Take the reciprocal on both side we get,
Rp1=20{R_{p1}} = 20

Now we have to calculate the value of Rs2{R_{s2}} with the series combination of R5,R6,Rp1{R_5},{R_6},{R_{p1}}
Using the series combination formula, we get, Rs2=R5+R6+Rp1{R_{s2}} = {R_5} + {R_6} + {R_{p1}}

On putting the values of these resistances, we find that,
Rs2=10+10+20\Rightarrow {R_{s2}} = 10 + 10 + 20
Rs2=40\Rightarrow {R_{s2}} = 40
Now Rs2,R7{R_{s2}},{R_7} are in parallel combination, So we have calculate the value of Rp2{R_{p2}} by using the parallel combination formula that is,

1Rp2=1R7+1Rs2   \dfrac{1}{{{R_{p2}}}} = \dfrac{1}{{{R_7}}} + \dfrac{1}{{{R_{s2}}}} \\\ \\\

On putting these value, we get,
1Rp2=140+140 1Rp2=240 1Rp2=120  \dfrac{1}{{{R_{p2}}}} = \dfrac{1}{{40}} + \dfrac{1}{{40}} \\\ \Rightarrow \dfrac{1}{{{R_{p2}}}} = \dfrac{2}{{40}} \\\ \Rightarrow \dfrac{1}{{{R_{p2}}}} = \dfrac{1}{{20}} \\\

Taking the reciprocal on both side we get,
Rp2=20\therefore {R_{p2}} = 20

Now we have to calculate final the value of Rnet{R_{net}} which are in series combination with R8,R9,Rp2{R_8},{R_9},{R_{p2}}

On putting the values of these we get,
Rnet=10+10+20\Rightarrow {R_{net}} = 10 + 10 + 20
After calculation we will get,
Rnet=40\therefore {R_{net}} = 40

Therefore, option (A) is the correct option.

Note: As we know that if the resistance is in series, the net resistance of the circuit will increase or if the resistance is in parallel than the resistance of the circuit decreases. So, these combinations are used on the requirement.