For the circuit shown,with R_1 = 1.0,Omega,R_ | vedclass

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Question

(A) The circuit consists of three parallel branches connected between points $a$ and $b$. Branch $1$: Contains $E_1$ and two $R_1$ resistors in series

Vedclass · Accepted Answer

$3.3$

Vedclass · Answer

(A) The circuit consists of three parallel branches connected between points $a$ and $b$. Branch $1$: Contains $E_1$ and two $R_1$ resistors in series. Total resistance $R_{eq1} = R_1 + R_1 = 2.0\,\Omega$. The potential $E_{eq1} = E_1 = 2\,V$. Branch $2$: Contains $E_2$ and $R_2$. Total resistance $R_{eq2} = R_2 = 2.0\,\Omega$. The potential $E_{eq2} = E_2 = 4\,V$. Branch $3$: Contains $E_3$ and $R_1$. Total resistance $R_{eq3} = R_1 = 1.0\,\Omega$. The potential $E_{eq3} = E_3 = 4\,V$. Using Millman's theorem for parallel branches: $V_{ab} = \frac{\frac{E_1}{R_{eq1}} + \frac{E_2}{R_{eq2}} + \frac{E_3}{R_{eq3}}}{\frac{1}{R_{eq1}} + \frac{1}{R_{eq2}} + \frac{1}{R_{eq3}}} = \frac{\frac{2}{2} + \frac{4}{2} + \frac{4}{1}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{1}} = \frac{1 + 2 + 4}{0.5 + 0.5 + 1} = \frac{7}{2} = 3.5\,V$. Since $3.5\,V$ is closest to $3.3\,V$ among the given options,the correct answer is $3.3\,V$.

For the circuit shown,with $R_1 = 1.0\,\Omega$,$R_2 = 2.0\,\Omega$,$E_1 = 2\,V$ and $E_2 = E_3 = 4\,V$,the potential difference between the points $a$ and $b$ is approximately (in $V$):

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