Interchanging Series Elements

Interchanging the elements in a series circuit does not effect the flow of current, the applied voltage and the consumption of power. All of the sources and loads will have similar behavior after replacing each other. Series elements may be the voltage sources, current sources, resistors or other passive elements (capacitors, inductors). It can be easily explained by the following example.
Example 1: Determine the voltage across resistor R1 and current through the circuit as shown in Fig.no.1. The circuit is redrawn in Fig.no.2. Where the voltage sources and resistors are replaced with each other.
Fig.no.1:
Solution: $$ R_T = R_1 + R_2$$ $$ R_T = 2 Ω + 5 Ω = 7 Ω$$ Applying KVL: $$E1-V_{R1} - E2 - V_{R2} = 0$$ $$ E1 - E2 = V_{R1}+ V_{R2}$$ $$ E1 - E2 = I R_1+ IR_2$$ $$ E1 - E2 = I( R_1+ R_2) = I R_T$$ $$ I = {E1 - E2 \over R_T}$$ $$ I = {10 V - 15 V \over 7Ω} = {-5 V \over 7 Ω}$$ $$ I = -3.57 A$$ The current with minus sign indicates that the direction of current is opposite to that of what we have shown in the circuit of Fig.no.1. If it was positive then our suggested current direction would be correct.
$$ V_1 = IR_1 = -3.57 \times 2 Ω$$ $$ V_1 = -7.14 V$$ The -ive and +ive polarity of V1 is also opposite to that of shown in fig.no.1. The circuit is redrawn in the following fig.no.2 with series elements interchanged.
Fig.no.2
$$ -E1 + E2 = I( R_1+ R_2) = I R_T$$ $$ I = {-E1 + E2 \over R_T}$$ $$ I = {-15 V + 10 V \over 7Ω} = {-5 V \over 7 Ω}$$ $$ I = -3.57 A$$ so the same voltage drop values will be calculated in both the circuits.
It is concluded from the example that interchanging series elements does not effect the flow of current in the circuit, thereby no effect will occur in the voltage drops and power.