First slide

Kirchoff's Law

Question

In the circuit given here the points A, B and C are 70 V , zero and 10 V respectively. Then

Moderate

A

The point ‘D’ will be at a potential of 60 V
B

The point ‘D’ will be at a potential of 20 V
C

Currents in the paths AD , DB and DC are in the ratio 1 : 2 : 3
D

Currents in the paths AD , DB and DC are in the ratio 3 : 2 : 1

Solution

$V_{A} {-V}_{D} =i × 10 \Rightarrow {70 - V}_{D} =10i \to \{1\} V_{D} {-V}_{B} {=i}_{1} × 20 \Rightarrow V_{D} {- 0 =20i}_{1} \to \{2\} V_{D} {-V}_{C} = (i - i_{1}) × 30 \Rightarrow V_{D} - 1 0 =30 (i - i_{1}) \to \{3\} F r o m e q . (i) & (i i) 70 - 20 i_{1} = 1 0 i \Rightarrow i + 2 i_{1} = 7 \to {4} F r o m e q . (i i) & (i i i) 20 i_{1} - 10 = 3 0 (i - i_{1}) \Rightarrow 3 i - 5 i_{1} = -1 \to {5} S o l v i n g \dots \dots (4) a n d (5) i_{1} = 2 A, i = 3 A, i - i_{1} = 1 A h e n c e V_{D} = 20 i_{1} = 40 V {i : i}_{1} : ({i-i}_{1}) = 3 : 2 : 1$

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