Question-210729

Question Number 210729 by peter frank last updated on 18/Aug/24

Answered by Spillover last updated on 18/Aug/24

Answered by Spillover last updated on 18/Aug/24

R = 3 k Ω + 1 k Ω + 2 k Ω = 6 k Ω = 6000 Ω

V = I R

I = \frac{V}{R} I = \frac{24}{6000} = 4 \times 10^{- 3} A

I = I_{1} + I_{2}

4 \times 10^{- 3} A = I_{1} + I_{2} \dots (i)

l o o p I

24 - 2000 I - v_{a} + v_{b} = 0

24 - 2000 (I_{1} + I_{2}) - v_{a} + v_{b} = 0

v_{a} - v_{b} = 24 - 2000 I_{1} - 2000 I_{2} \dots (i i)

l o o p I I

v_{a} - 1000 I_{1} - 3000 I_{1} - v_{b} = 0

v_{a} - v_{b} = 4000 I_{1} \dots . . (i i i)

s o l v e s i m u l t a n e o u s (i i) & (i i i)

v_{a} - v_{b} = 24 - 2000 I_{1} - 2000 I_{2}

v_{a} - v_{b} = 4000 I_{1}

I_{1} = 4 \times 10^{- 3} A I_{2} = 0

v_{a} - v_{b} = 4000 \times 4 \times 10^{- 3} A = 16 V

v_{a} - v_{b} = 16 V

Commented by peter frank last updated on 18/Aug/24

thanks mr spillover

Answered by mr W last updated on 18/Aug/24

V_{a - b} = \frac{4}{6} \times 24 = 16 V

Commented by peter frank last updated on 18/Aug/24

thanks mr W