In a circut, two resistors of 100 oham and 80 ohm are connected in parallel. The parallel group is then connected in series with a 100 ohm resistor. What is the total resistance of the circuit?
The equivalent resistance of two resistors in parallel can be found by
$\displaystyle \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$
So for your two parallel resistors we get an equivalent resistance of
$\displaystyle \frac{1}{R_{eq}} = \frac{1}{100~\Omega} + \frac{1}{80~\Omega} = \frac{9}{400}~\Omega ^{-1}$
So
$\displaystyle R_{eq} = \frac{400}{9}~\Omega$
The equivalent resistance of two resistors in series can be found by
$\displaystyle R_{eq} = R_1 + R_2$
So we are connecting the $\displaystyle \frac{400}{9}~\Omega$ equivalent resistance in series with a $\displaystyle 100~\Omega$ resistor:
$\displaystyle R_{eq} = \left ( \frac{400}{9}~\Omega \right ) + (100~\Omega)$
$\displaystyle R_{eq} = 144.44~\Omega$
-Dan