Find the net emf of the three batteries shown in figure - 2V 192 HA 4V 192 192 6V (C) 2V (A) SV (B) 6V (D) 3V O а

In the uniform electric field shown in figure. The value (v_C-v_D)

In uniform electric field, E = 10 NC^(-1) as shown in figure, find (i) V_(A)-V_(B) (ii) V_(B)-V_(C) .

The two diodes A and B are biased as shown, then -5V A -9V -3V B -6V

In the uniform electric field shown in figure, find : a. V_A-V_D b. V_A-V_C c. V_B-V_D d. V_C-V_D

In the uniform electric field shown in figure, find : a. V_A-V_D b. V_A-V_C c. V_B-V_D d. V_C-V_D

A battery of e.m.f. 10 V is connected to resistance as shown in figure. The potential difference V_(A) - V_(B) between the point A and B is

A battery of e.m.f. 10 V is connected to resistance as shown in figure. The potential difference V_(A) - V_(B) between the point A and B is

If the above question, the potential has correct relations as given= A. V_(A) gt V_(B) gt V_(C) gt V_(D) B. V_(A) gt V_(B) ge V_(C) gt V_(D) C. V_(D)=V_(C)=V_(B)=V_(A) D. V_(C) lt V_(B) gt V_(A) gt V_(D)