CBSE Class 10 Answered

Please find equivalent resistance in each case

Asked by jackiechrs10 | 07 Feb, 2019, 11:19: PM

Expert Answer

Question No. (i) is answered below

Given network is redrawn as shown above to calculate equivalent resistance.

Let us asssume a potential difference of V volts applied across A and B. Current flowing in the circuit is shown in the figure.

Let us apply Kirchoff's law for the closed path A-P-X-Q-B-A

we have 30×i₁+45×(i₁+i₃) = V or 75×i₁+45×i₃ = V or 5×i₁+3×i₃ = (V/15) .......................(1)

Let us apply Kirchoff's law for the closed path A-R-Y-S-B-A

we have 20×i₂ + 50×(i₂ - i₃) = V or 7×i₂ - 5×i₃ = (V/10) ..........................(2)

Let us apply Kirchoff's law for the closed path P-X-Y-R-P

we have 30×i₁ - 50×i₃ - 20×i₂= 0 or i₂ = (3/2) i₁ - (5/2)i₃............................(3)

By substituting for i₂ using eqn.(3) in eqn.(2) and after simplification, we get 21×i₁ - 45×i₃ = (V/5)..................(4)

By solving eqns. (1) and (4), we get, i₁ = ( V/80 ) and i₃ = (V/720)

we get i₂ from eqn.(3) using the values of i₁ and i₃ . Hence i₂= (11/720)V

total current drawn due to the potential difference is i = i₁ + i₂ = (1/80)V + (11/720)V = V/36

hence equivalent resistance = V / ( i₁ + i₂ ) = 36Ω

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Question no. (ii) is answered below

In the given network as shown in figure, across C and D , a 7Ω resistance is connected parallel to a

series combination of 10Ω and 4Ω resistances. Equivalent resistance of resistances connected across C and D

can be calculated and this equivalent resistance is (14/3)Ω.

The modified network after replacing the given resistances across C and D with equivalenet resistance

is shown right side of figure. Equivalent resistance of the modified network can be calculated,

since 8Ω is parallel with the seires combination of 3Ω and (14/3)Ω. Hence equivalent resistance across A and B is 3.9Ω

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Question no.(iii) is similar to Question no. (i).

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Question no.(iv) is answered below

If a current of i enters at inlet point A, it is divided into three parts i₁ , i₂ and i₃ and passing through the resistances as shown in figure.

There is a symmetry of resistance connections at inlet point A and at outlet point B.

If a given current i is divided at inlet A into three parts i₁ , i₂ and i₃ , then at oulet point B also same currents i₁ , i₂ and i₃are combined to give

total current i . Hence there is no current flow through the resistance connected at C and D and it can be removed

for calculating equivalent resistance as shown in the right side of figure.

Equivalent resistance is parallel combination of 2Ω, 6Ω and 1Ω resistances.

Equivalent resistance =

begin mathsize 12px style fraction numerator 2 cross times 6 cross times 1 over denominator 2 cross times 6 space plus space 6 cross times 1 space plus space 1 cross times 2 end fraction space equals space 0.75 space capital omega end style

Answered by Thiyagarajan K | 10 Feb, 2019, 11:33: PM

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