# Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 6)

6.

What is the voltage drop across

*R*_{1}?Discussion:

21 comments Page 1 of 3.
Dinkalem Mesfin said:
6 months ago

Applying KCL at node ------>(1)

i1-i2-i3=0

then in equation 1 apply KVL

12-Va/8 - Va/37 - (Va-4)/90 = 0.

50024 - 11966Va/ 226440 = 0.

Va=4.18

Then to find the voltage drop across R1.

Remember that voltage in sires is sum up.

Vt = V1+V2+V3+......Vn.

Va = 4.18,

Va = VR1 + (-12),

VR1 = 12-4.18,

VR1 = 7.82A.

i1-i2-i3=0

then in equation 1 apply KVL

12-Va/8 - Va/37 - (Va-4)/90 = 0.

50024 - 11966Va/ 226440 = 0.

Va=4.18

Then to find the voltage drop across R1.

Remember that voltage in sires is sum up.

Vt = V1+V2+V3+......Vn.

Va = 4.18,

Va = VR1 + (-12),

VR1 = 12-4.18,

VR1 = 7.82A.

Rajendra said:
4 years ago

Assume middle point as Va,

Calculate va

Then Vs1-Va = Voltage across R1.

Answer is 7.82 correct.

Calculate va

Then Vs1-Va = Voltage across R1.

Answer is 7.82 correct.

Ron said:
6 years ago

Let VB the voltage drop across 68 resistor and let VA the node between 68 and 37.

Lets solve first the VA :

VA-12/68 + VA/37 + VA-4/90 =0.

VA = 4.18 V.

Using loop analysis :

12-VB-4.18 = 0.

VB = 7.819 V.

Lets solve first the VA :

VA-12/68 + VA/37 + VA-4/90 =0.

VA = 4.18 V.

Using loop analysis :

12-VB-4.18 = 0.

VB = 7.819 V.

Chhatrapal said:
6 years ago

68/68 + 37 * 12 = 7.77 near to answer.

Lahu said:
6 years ago

First Calculate R1 current.

Using v = ir,

V = voltage drop across R1.

Using v = ir,

V = voltage drop across R1.

ALFRED said:
7 years ago

(VA - 12/68) + (VA - 4/90) + (VA/37) = 0.

=> (1.324 VA - 15.88/90) + (VA - 4/90) + 2.432 (VA/90) = 0

=> 1.324 VA + VA + 2.432 VA = 15.9 + 4

=> 4.75 VA = 19.9

=> VA = 19.9 / 4.75

=> 4.19#

=> 12 - 4.19 = 7.81#

=> (1.324 VA - 15.88/90) + (VA - 4/90) + 2.432 (VA/90) = 0

=> 1.324 VA + VA + 2.432 VA = 15.9 + 4

=> 4.75 VA = 19.9

=> VA = 19.9 / 4.75

=> 4.19#

=> 12 - 4.19 = 7.81#

Dhatrak Sonali said:
7 years ago

For loop 1.

= - 105I1 + 37I2 = - 12.....1st equation.

For 2nd loop.

37I1 - 127I2 = 4.

I1 = 0.114 A.

I2 = 0.002 A.

Voltage drop across R1.

V = R1*I1.

= 68*0.114.

V = 7.752 V.

= - 105I1 + 37I2 = - 12.....1st equation.

For 2nd loop.

37I1 - 127I2 = 4.

I1 = 0.114 A.

I2 = 0.002 A.

Voltage drop across R1.

V = R1*I1.

= 68*0.114.

V = 7.752 V.

BHEEMAN said:
7 years ago

Please solve this step by step with an equation.

Krishna said:
7 years ago

Using mesh analysis we get two equations we get the current passing through 1st loop as 0. 114 A which is the current flowing through R1.

So according to ohm's law V = IR thus the voltage drop across R1 is 0.114*68 the answer is 7.75 V.

So according to ohm's law V = IR thus the voltage drop across R1 is 0.114*68 the answer is 7.75 V.

Sysya said:
7 years ago

I don't get anything. Explain briefly.

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