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Question: Find the current through the \( 6\Omega \) and \( 4\Omega \) resistances. ![](https://www.vedantu....

Find the current through the 6Ω6\Omega and 4Ω4\Omega resistances.

Explanation

Solution

To solve this question, consider the three meshes given in the circuit and find the currents in each mesh. Then, using the values of mesh currents, find the values of current through the resistances.

Complete step by step solution:
Let us consider the current in the three meshes as I1{I_1} , I2{I_2} and I3{I_3} respectively, as in the given diagram.

Applying KVL in the mesh containing I1{I_1} current, we have
10I1+100+4(I1−I3)+6(I1−I2)=010{I_1} + 100 + 4({I_1} - {I_3}) + 6({I_1} - {I_2}) = 0
20I1+100−4I3−6I2=020{I_1} + 100 - 4{I_3} - 6{I_2} = 0
On rearranging, we get
−20I1+6I2+4I3=100- 20{I_1} + 6{I_2} + 4{I_3} = 100
Dividing both sides by 22
−10I1+3I2+2I3=50- 10{I_1} + 3{I_2} + 2{I_3} = 50 …………….(1)
Now, applying KVL in the mesh containing I2{I_2}
6(I2−I1)+12(I2−I3)+7I2−50=06({I_2} - {I_1}) + 12({I_2} - {I_3}) + 7{I_2} - 50 = 0
−6I1+25I2−12I3=50- 6{I_1} + 25{I_2} - 12{I_3} = 50…………….(2)
Finally, applying KVL in the mesh containing I3{I_3}
4(I3−I1)+14I3+12(I3−I2)=04({I_3} - {I_1}) + 14{I_3} + 12({I_3} - {I_2}) = 0
−4I1−12I2+30I3=0- 4{I_1} - 12{I_2} + 30{I_3} = 0
Dividing both sides by −2- 2
2I1+6I2−15I3=02{I_1} + 6{I_2} - 15{I_3} = 0 …………….(3)
Subtracting (1) from (2), we have
−6I1+25I2−12I3−(−10I1+3I2+2I3)=50−50- 6{I_1} + 25{I_2} - 12{I_3} - ( - 10{I_1} + 3{I_2} + 2{I_3}) = 50 - 50
4I1+22I2−14I3=04{I_1} + 22{I_2} - 14{I_3} = 0
Dividing both sides by 22
2I1+11I2−7I3=02{I_1} + 11{I_2} - 7{I_3} = 0 …………….(4)
Applying cross-multiplication method in (3) and (4), we have
I1123=I2−16=I310=λ\dfrac{{{I_1}}}{{123}} = \dfrac{{{I_2}}}{{ - 16}} = \dfrac{{{I_3}}}{{10}} = \lambda , where λ\lambda is some real number
∴I1=123λ, I2=−16λ, I3=10λ\therefore {I_1} = 123\lambda ,{\text{ }}{I_2} = - 16\lambda ,{\text{ }}{I_3} = 10\lambda...............(5)
Substituting these in (1), we get
−10(123λ)+3(−16λ)+2(10λ)=50- 10(123\lambda ) + 3( - 16\lambda ) + 2(10\lambda ) = 50
−1258λ=50- 1258\lambda = 50
So, we get
λ=−501258\lambda = - \dfrac{{50}}{{1258}}
Putting this in (5)
I1=−61501258=−4.89A{I_1} = - \dfrac{{6150}}{{1258}} = - 4.89A , I2=8001258=0.64A{I_2} = \dfrac{{800}}{{1258}} = 0.64A , and I3=−5001258=−0.39A{I_3} = - \dfrac{{500}}{{1258}} = - 0.39A
Now, current through the 6 Ω6\,\Omega resistance
I6Ω=I2−I1{I_{6\Omega }} = {I_2} - {I_1}
Putting the values, we get
I6Ω=0.64−(−4.89){I_{6\Omega }} = 0.64 - ( - 4.89)
I6Ω=5.53A{I_{6\Omega }} = 5.53A
Also, the current through the 4Ω4\Omega resistance
I4Ω=I3−I1{I_{4\Omega }} = {I_3} - {I_1}
Putting the values
I4Ω=−0.39−(−4.89){I_{4\Omega }} = - 0.39 - ( - 4.89)
I4Ω=4.5A{I_{4\Omega }} = 4.5A
Hence I6Ω=5.53A{I_{6\Omega }} = 5.53A and I4Ω=4.5A{I_{4\Omega }} = 4.5A .

Note:
While using KVL, first select a sign convention and then proceed. Do not get confused due to the sign convention. The choice of sign convention doesn’t change the final answer. We can use any one of the two sign conventions for the resistances and the batteries.