Question

Q1

lov a nov Hilr Tor 202 3204 sion b JOV TiH a] what is the current through the lo vemf near bottom left ? b] What is the potentiel difference Va-Vo ? the

Answer 1

* I 20 + R 20 ox lov с # I IZ IL کمه Ans: In the ciscuit diagsamo two 20.2 desistances are connected in parallel Let R be their equivalent desistance then [ By ķ=k+ķ] ko - to R=102 Subtiting this value of R=lose the equivalent circuit takes the Following form- tov d IZ SI lors e Iz 1 tov Il tov Past® With the Cussent distribution as shown above - Applying KCL at point e- I = Itt Ig Applying KVL in loop afeca To It TOV + 10(I +Iz) - 10/= 0. 21; +Ig in loop cabec lo Ig - 10V +10 (IL +Iz) - 10V = 0 If +212 = 2 V Now By solving, equebroms 0 and @ calculate I, and Iz Let us subtract equation @ from equation 0 b f e :av Applying kvi we can

Continued 211 + I2 = 21 -It - 21% = -2 I = 2 A IL -I, = 0 I = I2 Now flom equation 0 - 21; +I=8 3 I = 2 = .667 A from the diagram, we see that I, is the Cussent through 10V EME KAGGED neas the bottom left Hence- Cussent through desised Lo V EME = .667 A Anc Past b): Va - Vbo Va-Vc) + L Ve-kd) + (do Vb) [ From the circuit diagsom ] Va-Vb = Tov - 10 V + 10 I2 Va-Vb = 10xV [: I = I2 = 242 Ve-Vb = 2 Y = 6.678 Hence- Va-Vb = 6.67V Ans
Problem is solved using the concept of parallel combination of resistance, KCL, KCL and ohm's law.