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I am not looking for calculations, I can do that. Please forgive the drawing, my picture wont upload. However, (R1 and R2) are in series. Likewise, (R3 and R4) are also in series. I could not show the connections,but (R1 and R2) || (R3 and R4). I am just confused in how to treat this circuit. I already got an equivalent circuit to get the total current and total voltage. However, because there are no voltage drops across parallel resistors, but only series resistors, would that mean that there wouldn't be a voltage drop across the series resistors (R1 and R2) and (R3 and R4)?

                          ___R1___/\/\/\/\_____R2_/\/\/\/\__________

                          ____R3_/\/\/\/\______R4_/\/\/\/\__________
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    $\begingroup$ There ARE voltage drops across parallel resistors. Each parallel resistor has the same voltage drop across it. $\endgroup$ Mar 17 '20 at 17:26
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Ohm's law apply. The voltage drop across the $i$-the resistor is given by $V_i = R_i I_i$. Hence, in your case we have \begin{align} I_1 &= I_2,\\ I_3 &= I_4, \end{align} as well as \begin{align} V_{12} &= I_1 R_1 + I_2 R_2 \quad = \quad V_{34} = I_3 R_3 + I_4 R_4 \end{align}

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  • $\begingroup$ Perfect! I did this and the voltages summed to the source voltage. So it worked. Thanks! $\endgroup$ Mar 17 '20 at 17:42
  • $\begingroup$ Please read about Kirchhoff's law. $\endgroup$
    – Semoi
    Mar 17 '20 at 17:52

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