The current flowing through the 1 Omega resis | vedclass

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(B) Let the potential at node $C$ be $y$ and the potential at node $A$ be $x$. The potential at the junction between the $10 \ V$ battery and $1 \Omeg

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Let the potential at node $C$ be $y$ and the potential at node $A$ be $x$. The potential at the junction between the $10 \ V$ battery and $1 \Omega$ resistor is $(x-10) \ V$. Applying nodal analysis at node $C$ (potential $y$): $\frac{y-5}{2} + \frac{y-0}{2} + \frac{y-(x-10)}{1} = 0$ $y-5 + y + 2y - 2x + 20 = 0$ $4y - 2x + 15 = 0 \quad \dots(i)$ Applying nodal analysis at node $A$ (potential $x$): $\frac{x-5}{4} + \frac{x-0}{4} + \frac{x-10-y}{1} = 0$ $x-5 + x + 4x - 40 - 4y = 0$ $6x - 4y - 45 = 0 \quad \dots(ii)$ Adding $(i)$ and $(ii)$: $(4y - 2x + 15) + (6x - 4y - 45) = 0$ $4x - 30 = 0 \implies x = 7.5 \ V$ Substituting $x = 7.5$ in $(i)$: $4y - 2(7.5) + 15 = 0 \implies 4y - 15 + 15 = 0 \implies y = 0 \ V$ The current $I_1$ through the $1 \Omega$ resistor is given by: $I_1 = \frac{y - (x-10)}{1} = \frac{0 - (7.5 - 10)}{1} = \frac{2.5}{1} = 2.5 \ A$ Given $I_1 = \frac{n}{10} \ A$,we have $\frac{n}{10} = 2.5 \implies n = 25$.

The current flowing through the $1 \Omega$ resistor is $\frac{n}{10} \ A$. The value of $n$ is. . . . . . .

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