find the voltage across each resistor using kirchhoff rules

### Asked by carthi66 | 11th Nov, 2021, 02:49: PM

Expert Answer:

###
Current distribution in the circuit is assumed as shown in figure.
If we apply Kirchoff's voltage law to the closed loop ADCEFA , we get the following equation
4 i_{2} + 3 ( i_{2} + i_{3} ) + ( i_{1} + i_{2} ) = 4 ..............................(1)

If we apply Kirchoff's voltage law to the closed loop ABCEFA , we get the following equation
i_{1} + 2 ( i_{1} - i_{3} ) + ( i_{1} + i_{2} ) = 4 ..............................(2)
By subtracting eqn.(2) from eqn.(1) , we get
7 i_{2} - 3 i_{1} + 5 i_{3} = 0 or 7 i_{2} + 5 i_{3} = 3 i_{1} ......................... (3)
If we apply Kirchoff's voltage law to the closed loop BDCB , we get the following equation
5 i_{3} + 3 ( i_{2} + i_{3} ) + 2 ( i_{1} - i_{3} ) = 0
By simplifying above expression we get , 3 i_{2} + 10 i_{3} = 2 i_{1} ......................(4)
By solving eqn.(3) and (4) , we get i_{2} = (4/11) i_{1} and i_{3} = (1/11) i_{1}
By substituting i_{2} and i_{3} in eqn.(1) , we get i_{1} = (22/23) A
Hence i_{2} = (4/11) × (22/23) A = ( 8 / 23 ) A and i_{3} = ( 1/11) × (22/23) A = ( 2 / 23 ) A
Voltage drop across 1Ω resistor between A and B = 1 × ( 22/23) = (22 / 23 ) V = 0.956 V
Voltage drop across 2Ω resistor between B and C = 2 × [ ( 22/23) - (2/23) ] = (40 / 23 ) V = 1.739 V
Voltage drop across 3Ω resistor between D and C = 3 × [ ( 8/23) + (2/23) ] = (30 / 23 ) V = 1.304 V
Voltage drop across 4Ω resistor between A and D = 4 × (8/23) ] = (32 / 23 ) V = 1.391 V
Voltage drop across 5Ω resistor between B and D = 5 × (2/23) = (10 / 23 ) V = 0.435 V
Voltage drop across 1Ω resistor between E and F = 1 × [ ( 22/23) + (8/23) ] = (30 / 23 ) V = 1.304 V

_{2}+ 3 ( i

_{2}+ i

_{3}) + ( i

_{1}+ i

_{2}) = 4 ..............................(1)

_{1}+ 2 ( i

_{1}- i

_{3}) + ( i

_{1}+ i

_{2}) = 4 ..............................(2)

_{2}- 3 i

_{1}+ 5 i

_{3}= 0 or 7 i

_{2}+ 5 i

_{3}= 3 i

_{1}......................... (3)

_{3}+ 3 ( i

_{2}+ i

_{3}) + 2 ( i

_{1}- i

_{3}) = 0

_{2}+ 10 i

_{3}= 2 i

_{1}......................(4)

_{2}= (4/11) i

_{1}and i

_{3}= (1/11) i

_{1}

_{2}and i

_{3}in eqn.(1) , we get i

_{1}= (22/23) A

_{2}= (4/11) × (22/23) A = ( 8 / 23 ) A and i

_{3}= ( 1/11) × (22/23) A = ( 2 / 23 ) A

### Answered by Thiyagarajan K | 12th Nov, 2021, 12:26: AM

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