Find the current in the loop.

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(D) In the given circuit, there are three batteries. Let us assume the current $I$ flows in the clockwise direction.The electromotive forces (EMFs) ar

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(D) In the given circuit, there are three batteries. Let us assume the current $I$ flows in the clockwise direction.The electromotive forces (EMFs) are $40\, V$, $10\, V$, $15\, V$, and $20\, V$. Looking at the polarities:$40\, V$ and $15\, V$ batteries are pushing current in the clockwise direction, while $10\, V$ and $20\, V$ batteries are pushing in the counter-clockwise direction.Net $EMF$ $E_{net} = (40 + 15) - (10 + 20) = 55 - 30 = 25\, V$.The total resistance $R_{net}$ is the sum of all internal and external resistances:$R_{net} = 1\, \Omega + 2\, \Omega + 2\, \Omega + 1\, \Omega + 1\, \Omega + 3\, \Omega + 1\, \Omega = 11\, \Omega$.Using Ohm's law for the loop, $I = \frac{E_{net}}{R_{net}} = \frac{25}{11}\, A$.

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