The resistances $R_{1}$ and $R_{2}$ are connected in parallel. The equivalent resistance $R_{p}$ of the combination is:

(a) $R_{1} + R_{2}$

(b) $R_{1} - R_{2}$

(c) $\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

(D) $\frac{R_{1} + R_{2}}{R_{1} R_{2}}$

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Solution

Let $R_{p}$ be the equivalent resistance of the parallel combination of the resistor.
According to Ohm's law :

$I = I_{1} + I_{2}$
$\frac{V}{R_{p}} = \frac{V}{R_{1}} + \frac{V}{R_{2}}$
$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{p}} = \frac{R_{1} + R_{2}}{R_{1} R_{2}}$

$R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$

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The resistances R1 and R2 are connected in parallel. The equivalent resistance Rp of the combination is: (a) R1+R2 (b) R1−R2 (c) R1R2R1+R2 (D) R1+R2R1R2

Let Rp be the equivalent resistance of the parallel combination of the resistor. According to Ohm's law : I=I1+I2 VRp=VR1+VR2 1Rp=1R1+1R2 1Rp=R1+R2R1R2 RP=R1R2R1+R2

The resistances $R_{1}$ and $R_{2}$ are connected in parallel. The equivalent resistance $R_{p}$ of the combination is:

(a) $R_{1} + R_{2}$

(b) $R_{1} - R_{2}$

(c) $\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

(D) $\frac{R_{1} + R_{2}}{R_{1} R_{2}}$