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 R1?
Discussion:
21 comments Page 1 of 3.
Dinkalem Mesfin said:
3 years 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.
(2)
Nabeel Hayat said:
1 decade ago
There are two loop, You have to find Loop 1 Current, using KVL, but as on resistor 37Ohm both current I1 and I2 are passing, so we also have to find Current I2,
For that we have to use KVL on both the loops, we have two equations, from that You can find I1,
After that,
V=IR.
V=I (68).
For that we have to use KVL on both the loops, we have two equations, from that You can find I1,
After that,
V=IR.
V=I (68).
Krishna said:
10 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.
ALFRED said:
9 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#
Siddhartha said:
1 decade ago
Applying super position theorems and by solving them expression can be obtained i.e.,
((x-12)/68)=((x-14)/90).
On solving above expression we get x value as 3.67v.
Voltage drop is v1-x
i.e., 12-3.67=8.33v.
((x-12)/68)=((x-14)/90).
On solving above expression we get x value as 3.67v.
Voltage drop is v1-x
i.e., 12-3.67=8.33v.
Ron said:
8 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.
Dhatrak Sonali said:
10 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.
Nouman said:
1 decade ago
First Calculate IR1:
IR1 is 0.1148 A.
R1 = 68 ohms.
Vr1 = Ir1 x R1 = 0.1148 x 68 = 7.8 V.
The answer is 7.8 V.
IR1 is 0.1148 A.
R1 = 68 ohms.
Vr1 = Ir1 x R1 = 0.1148 x 68 = 7.8 V.
The answer is 7.8 V.
Rajendra said:
7 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.
Valentin said:
1 decade ago
The correct answer is 4,18~4.19V. To solve this problem is used the overlapping effects.
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