Answer
1. All in series
$R = 5100\Omega$
2. All in parallel
$R = 566.7\Omega$
3. One in parallel and two in series
$R = 1133\Omega$
4. One in series and two parallel
$R = 3988\Omega$
Work Step by Step
1. All in series:
In series circuits, the resistors add up so:
$\sum R = 1700 + 1700 + 1700 = 5100\Omega$
2. All parallel:
In parallel, the formula goes: $\frac{1}{\sum R} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3}$
So that becomes $\frac{1}{\sum R} = \frac{1}{1700} +\frac{1}{1700} + \frac{1}{1700}$
$\sum R = 566.7\Omega$
3. One in parallel and two in series:
First add the two series circuits:
$R = 1700 + 1700 = 3400\Omega$
Then use $\frac{1}{\sum R} = \frac{1}{R_1} +\frac{1}{R_2}$;
$\frac{1}{\sum R} = \frac{1}{1700} +\frac{1}{3400}$
$\sum R = 1133\Omega$
4. One in series and two in parallel
First add the two parallel:
$\frac{1}{R} = \frac{1}{R_1} +\frac{1}{R_2}$
$\frac{1}{R} = \frac{1}{1700} +\frac{1}{1700}$
$R = 2288\Omega$
Then add the result from the parallel and the one in series.
$\sum R = 2288 + 1700 = 3988\Omega$
