When we are calculating the resistance of a circuit in parallel, why do we take the formula as reciprocal of 1/R1+1/R2, and for the series circuit, we take it as R1+R2?

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These are the formulas for equivalent resistance: the equivalent resistance of the series circuit with the resistors `R_1` and `R_2` is `R_(eq) = R_1 + R_2` , and the equivalent resistance of the parallel circuit is determined by the formula

`1/R_(eq) = 1/R_1 + 1/R_2`

These formulas come from the...

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These are the formulas for equivalent resistance: the equivalent resistance of the series circuit with the resistors `R_1` and `R_2` is `R_(eq) = R_1 + R_2` , and the equivalent resistance of the parallel circuit is determined by the formula

`1/R_(eq) = 1/R_1 + 1/R_2`

These formulas come from the way the current flows through the circuits, and are easy to derive:

For the series circuit, the same current must flow through both resistors as the current that flows through the battery. This comes from the law of conservation of charge (current is the rate of the flow of charge.) So, the potential difference supplied by the battery will be split between the two resistors. According to the Ohm's Law, each potential difference is proportional to the resistance:

`U = IR_(eq)` (potential difference supplied by the battery)

`U_1 = IR_1` (potential difference on the first resistor)

`U_2 = IR_2` (potential difference on the second resistor)

`U = U_1 + U_2`

`IR_(eq) = IR_1 + IR_2`

Canceling out the current results in the formula for equivalent resistance of the series circuit: `R_(eq) = R_1 + R_2` .

In the parallel circuit, the current flowing in the circuit will split between the two resistors: `I = I_1 + I_2` . The potential difference, however, must be the same on each resistor and equal to the potential difference supplied by the battery, because the potential difference between the two points does not depend on the path between the two points. Again, the potential difference can be determined from the Ohm's Law:

`U = IR_(eq) = I_1R_1 = I_2R_2`

Expressing currents from here and plugging them in `I = I_1 + I_2` , we find

`U/R_(eq) = U/R_1 + U/R_2`

Canceling out U results in the formula for the equivalent resistance of the parallel circuit:

`1/R_(eq) = 1/R_1 + 1/R_2`.

 

 

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