Electrical Engineering - Branch, Loop and Node Analyses - Discussion
Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 2)
2.
What is the current through R2?
Discussion:
16 comments Page 1 of 2.
Harpreet said:
1 decade ago
Simply apply node base analysis and solve it by KCL Law.
Samyu said:
1 decade ago
Apply nodal analysis find voltage Va then by applying ohms law.
Ooops said:
1 decade ago
Isn't there any simple way to solve that porblem.
Rita J said:
1 decade ago
Applying kcl at node A and taking node B as refrence:
(15-va)/68-va/30-(va-8)/100=0
va=5.1689v;
Ir2=va/30=5.1689/30=0.172A=172mA;
(15-va)/68-va/30-(va-8)/100=0
va=5.1689v;
Ir2=va/30=5.1689/30=0.172A=172mA;
Ashish singhal said:
1 decade ago
By appling mesh analysis
15=68i'+ 30(i'-i")
15=98i'-30i"
Now at second battery
-8=100i"+30*(i'-i")
-8=-30i'+130i"
After calculation
i'=.145
i"= -.028
So total current in 30 ohm resistance
i' - i"
0.145 - (-0.028)
= 0.173 A or
= 173 mA.
15=68i'+ 30(i'-i")
15=98i'-30i"
Now at second battery
-8=100i"+30*(i'-i")
-8=-30i'+130i"
After calculation
i'=.145
i"= -.028
So total current in 30 ohm resistance
i' - i"
0.145 - (-0.028)
= 0.173 A or
= 173 mA.
Siddhartha said:
1 decade ago
Applying kvl to the circuit,
((v-15)/68)+(v/30)+((v-8)/100).
On solving above expression we get 5.171v.
Current at R2= 5.17/30=0.172 amps.
172 milli amps.
((v-15)/68)+(v/30)+((v-8)/100).
On solving above expression we get 5.171v.
Current at R2= 5.17/30=0.172 amps.
172 milli amps.
Dinesh kumar t said:
1 decade ago
Applying kcl at node A and taking node B as refrence:
(15-va)/68-va/30-(va-8)/100=0
va=5.1689v;
Ir2=va/30=5.1689/30=0.172A=172mA.
(15-va)/68-va/30-(va-8)/100=0
va=5.1689v;
Ir2=va/30=5.1689/30=0.172A=172mA.
Sunilkumar v said:
10 years ago
Apply nodal analysis for the circuit. After getting Va value, we can find current at 30 ohm.res
(15-VA)/68 - (VA-0)/30 - (VA-8)/100 = 0.
VA = 5.169.
R2 = VA/30.
R2 = 5.169/30.
R2 = 173 mA.
(15-VA)/68 - (VA-0)/30 - (VA-8)/100 = 0.
VA = 5.169.
R2 = VA/30.
R2 = 5.169/30.
R2 = 173 mA.
Julius said:
8 years ago
Can you please elaborate the solution so that I can understand. Thanks.
Pente vinodkumar said:
8 years ago
By applying loop analysis at first loop.
15 = 68i'+ 30(i'+i").
15 = 98i'+ 30i".
Now at second loop:
8 = 100i"+30*(i'+i").
8 = 30i'+130i".
After calculation:
i' = 0.145.
i" = 0.028.
So total current in 30 ohm resistance.
i' + i".
0.145 + 0.028.
= 173 mA.
15 = 68i'+ 30(i'+i").
15 = 98i'+ 30i".
Now at second loop:
8 = 100i"+30*(i'+i").
8 = 30i'+130i".
After calculation:
i' = 0.145.
i" = 0.028.
So total current in 30 ohm resistance.
i' + i".
0.145 + 0.028.
= 173 mA.
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