Let's solve each part step-by-step:
### (a) Series Connection:
When resistors are connected in series, the effective resistance [tex]\( R_{\text{series}} \)[/tex] is the sum of all individual resistances. The formula for the total resistance in series is:
[tex]\[ R_{\text{series}} = R_1 + R_2 + R_3 \][/tex]
Given the resistances:
[tex]\[ R_1 = 1 \, \Omega \][/tex]
[tex]\[ R_2 = 2 \, \Omega \][/tex]
[tex]\[ R_3 = 3 \, \Omega \][/tex]
Plugging in the values:
[tex]\[ R_{\text{series}} = 1 \Omega + 2 \Omega + 3 \Omega \][/tex]
[tex]\[ R_{\text{series}} = 6 \Omega \][/tex]
So, the effective resistance in series is [tex]\( 6 \Omega \)[/tex].
### (b) Parallel Connection:
When resistors are connected in parallel, the reciprocal of the effective resistance [tex]\( R_{\text{parallel}} \)[/tex] is the sum of the reciprocals of the individual resistances. The formula for the total resistance in parallel is:
[tex]\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \][/tex]
Given the resistances:
[tex]\[ R_1 = 1 \, \Omega \][/tex]
[tex]\[ R_2 = 2 \, \Omega \][/tex]
[tex]\[ R_3 = 3 \, \Omega \][/tex]
Plugging in the values:
[tex]\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{1 \, \Omega} + \frac{1}{2 \, \Omega} + \frac{1}{3 \, \Omega} \][/tex]
[tex]\[ \frac{1}{R_{\text{parallel}}} = 1 + 0.5 + 0.333 \][/tex]
[tex]\[ \frac{1}{R_{\text{parallel}}} = 1.833 \][/tex]
The effective resistance is calculated as:
[tex]\[ R_{\text{parallel}} = \frac{1}{1.833} \][/tex]
[tex]\[ R_{\text{parallel}} \approx 0.545 \, \Omega \][/tex]
So, the effective resistance in parallel is approximately [tex]\( 0.545 \, \Omega \)[/tex].
### Summary:
- In series: The effective resistance is [tex]\( 6 \Omega \)[/tex].
- In parallel: The effective resistance is approximately [tex]\( 0.545 \, \Omega \)[/tex].
