The currents i_1 and i_2 through the resistor | vedclass

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Question

(A) To find the currents,we apply Kirchhoff's laws or nodal analysis to the circuit.Let the potential at the junction connecting $E_2$ and $R_2$ be $V

Vedclass · Accepted Answer

$0.2\, A, 0.1\, A$

Vedclass · Answer

(A) To find the currents,we apply Kirchhoff's laws or nodal analysis to the circuit.Let the potential at the junction connecting $E_2$ and $R_2$ be $V = 3\,V$ (as shown in the reference solution diagram).For the branch containing $R_1$,the potential difference across the resistor is $V_{R1} = 3\,V - 1\,V = 2\,V$.Thus,the current $i_1$ is given by $i_1 = \frac{V_{R1}}{R_1} = \frac{2\,V}{10\,\Omega} = 0.2\,A$.For the branch containing $R_2$,the potential difference across the resistor is $V_{R2} = 3\,V - 0\,V = 3\,V$.Thus,the current $i_2$ is given by $i_2 = \frac{V_{R2}}{R_2} = \frac{3\,V}{30\,\Omega} = 0.1\,A$.Therefore,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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