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 2 of 2.
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.
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.
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;
Ooops said:
1 decade ago
Isn't there any simple way to solve that porblem.
Samyu said:
1 decade ago
Apply nodal analysis find voltage Va then by applying ohms law.
Harpreet said:
1 decade ago
Simply apply node base analysis and solve it by KCL Law.
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