For the given circuit,V_A - V_C = text{ . . . | vedclass

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(A) To find $V_A - V_C$,we travel from point $A$ to point $C$ through point $B$.Starting from $A$,the current $I = 3 \ A$ flows towards $B$ through th

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) To find $V_A - V_C$,we travel from point $A$ to point $C$ through point $B$.Starting from $A$,the current $I = 3 \ A$ flows towards $B$ through the $4 \ \Omega$ resistor.Applying the potential drop formula $V_B = V_A - I R + E$,we move from $A$ to $B$:$V_B = V_A - (3 \ A)(4 \ \Omega) + 5 \ V = V_A - 12 + 5 = V_A - 7$.Now,moving from $B$ to $C$ through the $3 \ V$ battery and $10 \ \Omega$ resistor,the current $I = 3 \ A$ flows from $B$ to $C$:$V_C = V_B - E - I R = V_B - 3 - (3 \ A)(10 \ \Omega) = V_B - 3 - 30 = V_B - 33$.Substituting $V_B = V_A - 7$ into the equation for $V_C$:$V_C = (V_A - 7) - 33 = V_A - 40$.Therefore,$V_A - V_C = 40 \ V$.

For the given circuit,$V_A - V_C = \text{ . . . . . . } \ V$.

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