Find the current in the three resistors as sh | vedclass

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(B) Let the potential at point $A$ be $V_A = 0 \ V$. Moving from $A$ to $C$ through the $2 \ V$ battery,the potential at $C$ is $V_C = 0 + 2 = 2 \ V$.

Vedclass · Accepted Answer

$i_1=0, i_2=0, i_3=0$

Vedclass · Answer

(B) Let the potential at point $A$ be $V_A = 0 \ V$. Moving from $A$ to $C$ through the $2 \ V$ battery,the potential at $C$ is $V_C = 0 + 2 = 2 \ V$. Moving from $A$ to $B$ is not direct,but looking at the bottom branch,if we assume the potential at $B$ is $0 \ V$,then moving from $B$ to $D$ through the $2 \ V$ battery gives $V_D = 0 + 2 = 2 \ V$. The potential difference across the first resistor $R$ is $V_{CD} = V_C - V_D = 2 \ V - 2 \ V = 0 \ V$. Thus,the current $i_1 = \frac{V_{CD}}{R} = 0$. Similarly,for the second branch,the potential at $E$ is $V_E = V_C + 2 - 2 = 2 \ V$ and the potential at $F$ is $V_F = V_D + 2 - 2 = 2 \ V$. Thus,$V_{EF} = 0 \ V$,so $i_2 = 0$. Finally,for the third branch,$V_G = V_E + 2 - 2 = 2 \ V$ and $V_H = V_F + 2 - 2 = 2 \ V$. Thus,$V_{GH} = 0 \ V$,so $i_3 = 0$. Therefore,$i_1 = i_2 = i_3 = 0$.

Find the current in the three resistors as shown in the following figure.

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