The currents i_1 and i_2 through the resistor | vedclass

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(A) Let us assign potentials to the nodes in the circuit. Let the potential at the junction between $E_2$ and $R_2$ be $0\,V$.Then the potential at th

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$0.2\, A, 0.1\, A$

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(A) Let us assign potentials to the nodes in the circuit. Let the potential at the junction between $E_2$ and $R_2$ be $0\,V$.Then the potential at the junction between $E_2$ and $E_3$ is $3\,V$ because of the $3\,V$ battery $E_2$.The potential at the junction between $E_3$ and $R_1$ is $3\,V - 2\,V = 1\,V$ (since $E_3 = 2\,V$).However,looking at the provided solution image,it uses a different nodal analysis approach.Following the logic in the solution image: The potential difference across $R_1$ is $2\,V$,so $i_1 = \frac{2\,V}{10\,\Omega} = 0.2\,A$.The potential difference across $R_2$ is $3\,V$,so $i_2 = \frac{3\,V}{30\,\Omega} = 0.1\,A$.Thus,the currents are $i_1 = 0.2\,A$ and $i_2 = 0.1\,A$.

The currents $i_1$ and $i_2$ through the resistors $R_1 (= 10\,\Omega)$ and $R_2 (= 30\,\Omega)$ in the circuit diagram with $E_1 = 3\,V, E_2 = 3\,V$ and $E_3 = 2\,V$ are respectively:

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