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Question: In the circuit shown below, calculate the net resistance of the circuit. ![](https://www.vedantu.c...

In the circuit shown below, calculate the net resistance of the circuit.

Explanation

Solution

In electrical circuits, resistance is the property of some materials which opposes the flow of current across them in the circuit and these types of materials are known as resistors. We will use the general formula of calculating net resistances when connected in series and when connected in parallel to each other.

Formula used:
When two resistors of resistance say R1(and)R2{R_1}(and){R_2} connected in series, then their net resistance can be calculated as Rseries=R1+R2{R_{series}} = {R_1} + {R_2} whereas when these two same resistors are connected in parallel combination then, we can calculate their net resistance as 1Rparallel=1R1+1R2\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}.

Complete step by step answer:
Let us first find the net resistance among the series combination of resistances 10Ω(and)15Ω10\Omega (and)15\Omega let’s say net resistance among this series combination is R10,15{R_{10,15}} now, using series formula Rseries=R1+R2{R_{series}} = {R_1} + {R_2} we have,
R10,15=10+15{R_{10,15}} = 10 + 15
R10,15=25Ω(i)\Rightarrow {R_{10,15}} = 25\Omega \to (i)
Now, again we have resistances of 20Ω(and)5Ω20\Omega (and)5\Omega are connected in series so using again series formula Rseries=R1+R2{R_{series}} = {R_1} + {R_2} , their net resistance can be written as,
R20,5=20+5{R_{20,5}} = 20 + 5
R20,5=25Ω(ii)\Rightarrow {R_{20,5}} = 25\Omega \to (ii)

Now, the net resistances of R10,15(and)R20,5{R_{10,15}}(and){R_{20,5}} both are connected in parallel combination, so using parallel combination formula 1Rparallel=1R1+1R2\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} we can find net resistance of circuits as:
1Rnet=1R10,15+1R20,5\dfrac{1}{{{R_{net}}}} = \dfrac{1}{{{R_{10,15}}}} + \dfrac{1}{{{R_{20,5}}}}
Put the values from the equations R10,15=25Ω(i){R_{10,15}} = 25\Omega \to (i) and R20,5=25Ω(ii){R_{20,5}} = 25\Omega \to (ii) we have,
1Rnet=125+125\dfrac{1}{{{R_{net}}}} = \dfrac{1}{{25}} + \dfrac{1}{{25}}
1Rnet=225\Rightarrow \dfrac{1}{{{R_{net}}}} = \dfrac{2}{{25}}
Rnet=12.5Ω\therefore {R_{net}} = 12.5\Omega

Hence, the net resistance of the given circuit diagram is Rnet=12.5Ω{R_{net}} = 12.5\Omega .

Note: It should be remembered that, when two resistances are connected in series combination, the current flowing across both resistances are same but potential difference is different. while in parallel combination, the potential difference across both resistances is same but different current flows across both the resistances.