In the adjoining diagram R-1-10Omega- R-2-20Omega- R-3-40 Omega- R-4-80Omega and V-A-5 V- V-B-10 V- V-C-20V- V-D-15 V- The current in the resistance R-1 will be0-4 A towards O0-4 A away from O0-6 A towards O0-6 A away from O

Question

Toppr Admin · Accepted Answer

Let the potential at the common point of all branches be V-Then- using nodal analysis-5-x2212-V10-V-x2212-1020-V-x2212-2040-V-x2212-1580V-13515-I-5-x2212-V10-x2212-0-4-xA0-ASince- negative sign is there- that means the current will be away from O-

In the adjoining diagram $R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 40 Ω, R_{4} = 80 Ω$ and $V_{A} = 5 V, V_{B} = 10 V, V_{C} = 20 V, V_{D} = 15 V$ . The current in the resistance $R_{1}$ will be

0.4 A towards O
0.4 A away from O
0.6 A towards O
0.6 A away from O

Let the potential at the common point of all branches be $V$ .
Then, using nodal analysis:
$\frac{5 - V}{10} = \frac{V - 10}{20} + \frac{V - 20}{40} + \frac{V - 15}{80} V = \frac{135}{15}, I = \frac{5 - V}{10} = - 0.4 A$
Since, negative sign is there, that means the current will be away from $O$ .

In the adjoining diagram R1=10Ω,R2=20Ω,R3=40Ω,R4=80Ω and VA=5 V,VB=10 V,VC=20V,VD=15 V. The current in the resistance R1 will be0.4 A towards O0.4 A away from O0.6 A towards O0.6 A away from O

Let the potential at the common point of all branches be V.Then, using nodal analysis:5−V10=V−1020+V−2040+V−1580V=13515,I=5−V10=−0.4 ASince, negative sign is there, that means the current will be away from O.

In the adjoining diagram $R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 40 Ω, R_{4} = 80 Ω$ and $V_{A} = 5 V, V_{B} = 10 V, V_{C} = 20 V, V_{D} = 15 V$ . The current in the resistance $R_{1}$ will be

0.4 A towards O
0.4 A away from O
0.6 A towards O
0.6 A away from O

Let the potential at the common point of all branches be $V$ .
Then, using nodal analysis:
$\frac{5 - V}{10} = \frac{V - 10}{20} + \frac{V - 20}{40} + \frac{V - 15}{80} V = \frac{135}{15}, I = \frac{5 - V}{10} = - 0.4 A$
Since, negative sign is there, that means the current will be away from $O$ .