Answer to Question #170576 in Electric Circuits for Innocent

Question #170576

If the resistor are in parallel,prove also that the equivalent resistance is R where

1/R=1/R1 +1/R2 +1/R3+...1/RN


1
Expert's answer
2021-03-16T11:38:23-0400

Explanations & Calculations


  • For the array of resistors, the total current by it can be written as follows as the total current is the sum of currents through each branch: resistor.

I=i1+i2+i3+i4+.....+in\qquad\qquad \begin{aligned} \small I&=\small i_1+i_2+i_3+i_4+.....+i_n \end{aligned}

  • As the potential drop across each resistor is equal in a parallel arrangement, individual currents from the above equation can be replaced by i=VRi=\frac{V}{R},

I=VR1+VR2+VR3+VR4+.....+VRn\qquad\qquad \begin{aligned} \small I&=\small \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}+\frac{V}{R_4}+.....+\frac{V}{R_n} \end{aligned}

  • An equivalent resistance R\small R to those array of resistors is such, that it conducts the same total current I\small I having the same potential drop V\small V across it.
  • Then,

I=VR\qquad\qquad \begin{aligned} \small I&=\small\frac{V}{R} \end{aligned}

  • Then replacing the total current I\small I from both the equations we get

VR=VR1+VR2+VR3+VR4+....+VRn1R=1R1+1R2+1R3+1R4+.......+1Rn(V)\qquad\qquad \begin{aligned} \small \frac{V}{R}&=\small \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}+\frac{V}{R_4}+....+\frac{V}{R_n}\\ \small \frac{1}{R}&=\small \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+.......+\frac{1}{R_n}\cdots(\,\cancel{\small V}\,) \end{aligned}



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